Hi, noaksga:
The short answer is that yes, only ncubes have the property of
packing together to form a larger regular "solid" (more properly,
regular polytope) for dimensions greater than 2. I'll assume it is
obvious that ncubes have this "selfsimilar dissection" property in
all dimensions, focusing attention on the elimination of all other
possibilities.
I will present various arguments below to deal with all specific
cases. As a way of organizing the material, I'll explain for each
dimension:
1. What the regular polytopes are.
2. Which ones can be used to "tile" the plane or "fill space".
3. How formulas for volume, hypervolume, etc. apply to each.
Because you've asked what "can be combined to form a larger regular
solid," I've assumed that overlapping is not allowed. Throughout I'll
exclude from consideration the "nonconvex" regular solids (and their
higher dimensional analogs), because these are selfintersecting.
[However see my Additional Interest link at the end for some Web pages
where these nonconvex regular "solids" are discussed.]
* * * * * * * * * * * * * * *
Just to cover all the bases, in two dimensions there are infinitely
many convex regular polygons, one similarity class for any number
greater than 2 of equal sides and equal angles. Which of these can be
used to dissect a larger regular polygon?
In two dimensions we can:
 Dissect an equilateral triangle into smaller equilateral triangles.
 Dissect a square into smaller squares.
 Dissect a regular hexagon into equilateral triangles.
and that's all for regular dissections of regular polygons.
To prove this one might want to separate those cases (a) where a
regular polygon can be dissected into selfsimilar polygons from those
(b) where a regular polygon can be dissected into congruent regular
polygons dissimilar to the larger one.
In particular cases (a) must involve regular polygons that can tile
the plane. For if one such can be dissected into smaller "copies" of
itself, then turning the construction outword we could recursively
enlarge upon the original figure until the entire plane becomes filled
with tightly packed copies of itself.
It is known that the only regular polygons which tile the plane per
case (a) are equilateral triangles, squares, and regular hexagons. I
won't belabor the point here, but it foreshadows an argument we will
find useful in higher dimensions.
In case (b) we would direct attention to how the interior angle of the
larger regular polygon could be subdivided into a whole number
multiple of an interior angle of a smaller regular polygon. Since
already with a square the interior angle is pi/2 or 90 degrees, no
multiple of this (greater than 1) can be accomodated by the
(nonreflexive) interior angle of another regular polygon.
Having said as much (and having much more to cover about the higher
dimensional cases), I leave further details of the dissection of
regular polygons by regular polygons to the proverbial "interested
reader".
* * * * * * * * * * * * * * *
In three dimensions there are five convex regular polyhedra, all
widely known since the times of the ancient Greeks as Platonic Solids
because Plato enumerates them in his Timaeus, embuing them with
mystical significance:
 regular tetrahedron
 cube
 regular octahedron
 regular dodecahedron
 regular icosahedron
[Since we deal here only with regular polyhedra and their higher
dimensional counterparts, I'll omit "regular" in association with the
specific geometric names from now on.]
Again it is helpful to know that of these, only the cube is "space
filling", in the sense of being able to tightly pack copies of a
regular polyhedron so that no gaps in space occur:
[SpaceFilling Polyhedron]
(Eric W. Weisstein, From MathWorldA Wolfram Web Resource)
http://mathworld.wolfram.com/SpaceFillingPolyhedron.html
"A spacefilling polyhedron is a polyhedron which can be used to
generate a tessellation of space. Although even Aristotle himself
proclaimed in his work On the Heavens that the tetrahedron fills
space, it in fact does not (Hilbert and CohnVossen 1999, p. 45). The
cube is the only Platonic solid possessing this property (Gardner
1984, pp. 183184)."
Since the cube is spacefilling but none of the other regular
polyhedra are, we immediately have that the cube cannot be dissected
into any of the others, for that would imply they also were
spacefilling. Likewise, as in our suggested argument about regular
polygons, none of the regular polyhedra except the cube can be
dissected into smaller selfsimilar figures, for that too would imply
they can fill space.
To advance one other sort of general argument, we require formulas for
the volumes of these regular polyhedra, which can be found here:
[The Platonic Solids  David A. Fontaine]
http://davidf.faricy.net/polyhedra/Platonic_Solids.html
Assuming edges of unit length, we have these values:
polyhedron volume dihedral angle
  
tetrahedron SQRT(2)/12 arccos(1/3)
cube 1 pi/2
octahedron SQRT(2)/3 arccos(1/3)
dodecahedron (15+7SQRT(5))/4 2arctan((1+SQRT(5))/2)
icosahedron 5(3+SQRT(5)/12 2arctan((3+SQRT(5))/2)
From these volume formulas we can deduce the impossibility of
dissection for most combinations. If a larger regular polyhedron is
to be dissected into a whole number of equal smaller regular
polyhedra, then the edge length of the larger polyhedron must be an
integer multiple of the edge length of the smaller polyhedra. It
follows that the respective volumes of the two regular polyhedra for
unit edge lengths must be commensurate (have a rational ratio).
But among the distinct types of regular polyhedra, only the
tetrahedron and octahedron have a ratio of volumes that is rational.
So having dealt with the cases (a) of a regular polyhedron dissected
into selfsimilar copies by means of the spacefilling argument, we
have most of the cases (b) of a regular polyhedron dissected into
copies of a different kind disposed of by these volume formulas. I
summarize what the results are in a table with "footnotes" for the
tetrahedron/octahedron combinations that require still a special
argument:
\  Larger polyhedron
\  tetrahedron cube octahedron dodecahedron icosahedron
\  (1) (2) (3) (4) (5)


S 
m (1)  no no no[A] no no
a 
l 
l 
e (2)  no yes no no no
r 

p 
o (3)  no[B] no no no no
l 
y 
h 
e (4)  no no no no no
d 
r 
o 
n (5)  no no no no no

[A] If one tries to fit 4 tetrahedra into the corners of the
octahedron, one finds that they leave a gap, ie. they will not fill
out the octahedron's "solid angle" at a vertex. Concretely, the
dihedral angle (between two faces) of a regular tetrahedron is
arccos(1/3) or approximately 70.53 degrees, and the dihedral angle of
a regular octahedron is arccos(1/3) or approximately 109.47 degrees.
Clearly one tetrahedron is too "narrow" to fit along the edge of an
octahedron, and two tetrahedra flush facetoface will be too "wide"
to fit there.
[B] It is also clear from the angular data just cited that a regular
octahedron cannot fit (edge down) within the dihedral angle of the
tetrahedron; its shape is too "blunt".
This completes our analysis of the "regular solids" cases.
* * * * * * * * * * * * * * *
The most interesting dimension is four, where there are six regular
polytopes. Five of are analogs of the regular threedimensional
figures, and one is unique:
 hypertetrahedron (or 4simplex)
 hypercube (or 4cube, aka "tesseract")
 hyperoctahedron (or 4D crosspolytope, with 16 tetrahedral faces)
 hyperdodecahedron (or 120cell, with 120 dodecahedral faces)
 hypericosahedron (or 600cell, with 600 tetrahedral faces)
 the "24cell" (with 24 octahedral faces)
While some familiarity with the regular solids could be expected, the
generalizations to higher dimensions are not widely taught. For a
starting point, one might enjoy the fruits of this person's labors to
visualize the unfamiliar:
[Polytopes]
http://home.inreach.com/rtowle/Polytopes/polytope.html
[Polytopes (continued)]
http://home.inreach.com/rtowle/Polytopes/Chapter2/Polytopes2.html
"The ability of equal hypercubes to closepack and fill an nspace is
significant. It is not at all a commonplace, for a regular polytope to
fill space. Consider the plane: we may tile it with regular triangles,
with squares, or with hexagons. By mixing different types of regular
polygon, we may obtain a few more "Archimedean" tilings; however, our
options are rather limited. There is no tiling of regular polygons
which includes a pentagon, for instance, not to speak of a heptagon
(7gon), or a 17gon.
"In three dimensions, we have the spacefilling of cubes, and that is
that. There are a few Archimedean spacefillings by mixtures of
regular and Archimedean polyhedra. For instance, one may fill 3space
with Platonic tetrahedra and octahedra.
"In four dimensions, surprisingly, there are three spacefillings by
regular polytopes (the hypercube, the cross polytope, and the 24cell
may all be used to fill the 4space). But in all higher spaces there
is only the hypercube."
* * * * * * * * * * * * * * *
It is especially interesting that in four dimensions the 4cube is
_not_ the only "spacefilling" regular polytope. It must share that
property with the 4dimensional crosspolytope (the analog of the
octahedron) and with the "24cell", socalled because it has 24
octahedral "faces".
It nevertheless remains true that only the 4cube can be dissected
into regular polytopes, and only by using 4cubes to do it.
These "polychoron" (a term for the 4dimensional "solids" proposed by
Norman Johnson) have "hypervolumes" which are nicely summarized on a
Web page at this same site linked earlier:
[Regular Convex FourDimensional Polytopes]
http://davidf.faricy.net/polyhedra/Polytopes.html
Note that instead of "dihedral angles" (angles between faces sharing
an edge), we now have "dichoral angles" (angles between regular
polyhedron sharing a face).
Assuming edges of unit length, we have these values:
polychoron hypervolume dichoral angle
  
4simplex SQRT(5)/96 arccos(1/4)
4cube 1 pi/2
crosspolytope 1/6 2pi/3
"120cell" 15(105+47SQRT(5))/4 4pi/5
"600cell" 25(2+SQRT(5))/4 2arctan((2+SQRT(5))SQRT(3))
"24cell" 2 2pi/3
With three of these polychoron able to "fill space" and three unable
to do so, and with irrational ratios of the (normalized) volumes for
all combinations except among those which "fill space" (the 4cube,
the crosspolytope (aka hyperoctahedron), and the 24cell), we can
again summarize the "dissection" results, using "footnotes" where more
argument is needed:
\  Larger polychoron
\  4simplex 4cube crosspolytope "120cell" "600cell" "24cell"
\  (1) (2) (3) (4) (5) (6)


S 
m (1)  no no no no no no
a 
l 
l 
e (2)  no yes no[C] no no no[D]
r 

p 
o (3)  no no[E] no[F] no no no[G]
l 
y 
c 
h (4)  no no no no no no
o 
r 
o 
n (5)  no no no no no no



(6)  no no[H] no[I] no no no[J]

Notes: See earlier results on dissection of regular polyhedra as needed.
[C] If the crosspolytope were dissected into 4cubes, its tetrahedral
hyperfaces would have to be dissected into cubes but cannot be.
[D] If the 24cell were dissected into 4cubes, its octahedral
hyperfaces would have to be dissected into cubes but cannot be.
[E] If the 4cube were dissected into crosspolytopes, its cubic
hyperfaces would have to be dissected into tetrahedra but cannot be.
[F] If the crosspolytope were selfsimilarly dissected, its tetrahedral
hyperfaces would have to be selfsimilarly dissected but cannot be.
[G] If the 24cell were dissected into crosspolytopes, its octahedral
hyperfaces would have to be dissected into tetrahedra but cannot be.
[H] If the 4cube were dissected into 24cells, its cubic hyperfaces
would have to be dissected into octahedra but cannot be.
[I] If the crosspolytope were dissected into 24cells, its tetrahedral
hyperfaces would have to be dissected into octahedra but cannot be.
[J] If the 24cell were selfsimilarly dissected, its octahedral
hyperfaces would have to be selfsimilarly dissected but cannot be.
This completes our analysis of the regular polychora cases.
* * * * * * * * * * * * * * *
All convex regular polytopes ("solids" in possibly higher dimensions)
are described here:
[Platonic Solids in All Dimensions  John Baez]
http://math.ucr.edu/home/baez/platonic.html
However in dimension five and higher, the regular polytopes reduce to
only three families:
 nsimplexes (generalizing the triangle/tetrahedron)
 ncubes
 ndimensional crosspolytope (2^n "faces" which are (n1)simplexes)
and we recall from a previously quoted link that of these, only the
ncube is "space filling".
For the (hyper)volume V_n of a regular nsimplex with unit length edges, see:
[Regular Polytopes]
http://www.math.rutgers.edu/~erowland/polytopes.html
V_n = SQRT(n+1)/(2^(n/2) * n!)
I could not find a ready online reference for the (hyper)volume U_n of
a regular ndimensional crosspolytope with unit length edges, so I
worked it out:
U_n = (2^(n/2))/n!
Of course the (hyper)volume of the ncube with unit length edges is 1.
We are set to polish off all the higher dimensional regular dissection cases:
\  Larger polytope
\  nsimplex ncube crosspolytope
\  (1) (2) (3)

S 
m 
a (1)  no no no[K]
l 
l 
e 
r 
(2)  no[L] yes no[M]
p 
o 
l 
y 
t 
o (3)  no[N] no no
p 
e 
Notes:
[K] The 2^n "faces" of the ndimensional crosspolytope are
(n1)simplexes, as are the faces of the nsimplex. A dissection of
the larger polytope then requires either that (1) we have one
nsimplex covering each face of the crosspolytope, or else that (2)
the faces of the crosspolytope are themselves "selfsimilarly"
dissected into (n1)simplexes. We have already ruled out the latter
possibility (for n1 > 2, so it applies here).
Nor is it possible to cover each "face" of the crosspolytope with an
equal "face" of a different nsimplex, because the collective size of
those 2^n nsimplexes is greater than that of the crosspolytope.
That is:
U_n = (2^(n/2)/n! < SQRT(n+1)*(2^(n/2)/n! = (2^n)*V_n
[L] Both nsimplex and ndimensional crosspolytope have hyperfaces
that are (n1)simplexes, and the (n1)simplex cannot be dissected by
the (n1)cubes that are hyperfaces of the ncube.
[M] Similar to [K], we would either have to cover each (n1)simplex
"face" by an (n1)simplex "face" of a crosspolytope, or else dissect
a regular (n1)simplex into smaller regular (n1)simplexes (which is
not possible for n1 > 2). But the former option is impossible as
well because a single crosspolytope is already bigger than the
nsimplex we're trying to dissect:
V_n = SQRT(n+1)/(2^(n/2) * n!) < (2^n)/(2^(n/2) * n!) = U_n
for all n > 0.
This completes our analysis of the higher dimensional regular polytope cases.
* * * * * * * * * * * * * * *
Additional Interest:
As promised here's a link to a discussion/visualization of some
selfintersecting "regular" polyhedra, otherwise excluded from my
analysis but mentioned in some the Web pages linked to above:
[The KeplerPoinsot Polyhedra]
http://www.georgehart.com/virtualpolyhedra/keplerpoinsotinfo.html
Such nonconvex constructions also have higher dimensional counterparts.
* * * * * * * * * * * * * * *
If Clarification for some part of my Answer would be helpful, please let me know.
regards, mathtalkga 