How many cubic feet of air will pass through an orphic 0.1875
inches in diameter in one minute?

Surely, this depends on the pressure differential, since if there were
none, there would be no flow.

I'm guessing too that by "orphic", you mean orifice, not some device
found in the Underworld.

You must specify the medium (eg. air, its temperature and
pressure-meaning density and humidity) and the pressure differential
as well as the orifice area and its flow coifficent.

A simple hole in a thin plate will have a coefficent of .6-.62 whereas
a hole with a rounded entry will have a coefficent of 1.
To simplify using "standard" air density of .075 pounds per cubic
foot, the formula goes as follows-

Square Root of (pressure differential/.075) Times 1096.7 Times the
flow coefficent Times the orifice area in SQUARE FEET = cubic feet per
min.

for example
your orifice is .1875 and I will assume it to be a square edged hole
in a thin plate with a coefficent of .61. Its area will be
PI*(.09375*.09375)=.294525 square inches. .294525/144=.002045313sq ft

pressure differential will be assumed to be 10 inches of water

1096.7*sqrt(10/.075= 12663.6 feet per minite air speed
the orifice area of .002045313*12663.6 = 25.9 theoretical cubic feet per minite
25.9 times the coefficent of .61 = 15.79 cubic feet per minite@ 10in
water pressure differential.

For other pressure differentials just replace the 10 with any other
pressure in inches of water and recalculate.