How much power does a crystal radio generate?
What determines the output?
Can crystal radios be constructed in series or parallel to increase
voltage and current?
What components need to be replicated? Does each need a separate antenna?
Is there a practical limit to the number of units that can be connected together?
Can I drive a lamp? an LED? a load (motor) from such a set-up?

You've essentially asked nine questions here. In order to get a good
answer, you may want to consider raising your price considerably.

Price raised 9x$2 = $18.00

The thing to remember here is that it isn't the crystal radio that is
generating the power; the power is being transmitted by some distant
broadcaster, and is being recieved by the antenna.   The crystal radio
-- the stuff attached to the antenna -- is just a way of filtering out
some frequencies from all the signals being recieved from the antenna
and converting that into audio.

This starts to make the rest of the questions easier to answer.   It
is the antennae that you really need to replicate to absorb more
energy; the conversion into whatever useful form you want (audio, or
DC to drive an LED) does not need to be replicated.    Realistically,
these antenna are going to need to be some sizeable fraction of a
wavelength away from each other, which is going to set the practical
number you can string together.  (Think of the distance between bars
in a television antenna people used to have on their roofs; you can't
make the antenna more compact just by moving the bars closer together,
as the antenna won't work in the same way).

This consideration, amongst others, is why people who have tried such
experiments in the past to use very high frequency radio waves, such
as microwaves, for power-transmission uses.   An example of flying a
model airplane without its own power source -- powered only by
microwaves beamed from the ground -- is shown here;
http://www.kurasc.kyoto-u.ac.jp/space-group/sps/milax-e.html
And another URL discusses the possibility of sending power from solar
sattellites to the surface of the Earth with microwaves -
http://www.space.com/businesstechnology/technology/solar_power_sats_011017-1.html

And it would not be a good idea to try the microwave thing with parts
from the kitchen -- you could cook yourself I think.

Mike Brinn makes an excellent point; don't mess with microwaves unless
you know what you're doing; your body will do an excellent job of
absorbing microwaves and converting the energy into heat (or worse).

If you're interested in more quantitative answers about how much power
can be transmitted this way, you need to know something about the
power being broadcast, and the geometry of the receiving antenna
array.

For a transmitter broadcasting isotropically at some average power
level P at some distance r away from you, there will be a transmitted
power density at your position of (P / (4 pi r^2)).   The amount of
energy your antenna can ideally absorb is usually described in terms
of an effective area (see for instance
http://farside.ph.utexas.edu/teaching/jk1/lectures/node83.html ).  For
a simple dipole antenna, receiving at a wavelength of lambda, the
effective area is about lambda^2/8; other antenna geometries will give
different effective areas, but that's a good baseline to consider
from.   This means that the power available in principle to drive a 
load is something like

0.01 (lambda/r)^2 P.

Now, some of this power will be lost in the antenna feed and in
conversion, so this is an upper limit.  On the other hand, if a
directional antenna is being used to transmit the power in a more
focussed way to the reciever, that will increase the amount of power
available to you.

Since the original context was crystal radios, let's imagine an AM
radio station, transmitting at about 25,000 kW at around 1,000 kHz
somewhere near your hometown (say within 30 km).   I don't know what
the effective area of one of those coil antennae are, but let's
imagine it's about the same as a dipole (although I assume it's
significantly smaller).   The wavelength of the AM signal is about 300
m, so we have something like

0.01 (300m/30000m)^2 25,000 kW = 0.025 W

Whoops, typo; that should be *either* 25,000 W or 25 kW, not 25,000 kW
(which would be a scarily strong broadcasting station!)  If it were
actually 25,000 kW, then the recieved power would be 25W, not 0.025 W.

I once read a tale. Farmer owns land around powerful radio
transmitter. Puts cables in his fields and gets free power. Presumably
it would work being very close to the source, but for every Watt he
extracted whoever owned the transmitter would have to put in an extra
Watt.