Say you were exerting the hard but not unfeasible power of 300W to travel at 20 mpg in a 100" gear on a bike with fullsize wheels.
power = force * velocity
Therefore we need to find the velocity at which the chain is being pulled forward relative to the wheel by the chain ring.
100" x pi = 314" ≃ 8m travelled along the ground for every revolution of the pedals. 20 mph is approximately 10 m/s so our pedals will be turning at 10/8 = 1.25 Hz, i.e. a cadence of 75 rpm.
A 100" gear is a chainring/sprocket ratio of roughly 11/3, or a 44 tooth chainring to a 12 tooth sprocket. Let's say our chainring has a radius of diameter of 20 cm. We therefore pull chain at a rate of (20 cm * pi * 1.25) metres every second.
0.2 * 3.14 * 1.25 = Vchain ≃ 0.8m/s
Remember that power = force * velocity, so force = power/velocity
Fchain = 300/0.8 = 375 Newtons of force on the driving part of your chain (the chain tension)
At the rear sprocket, the torque (the turning force) = force * perpendicular distance from the centre of rotation. Since we're using a 12 sprocket, let's call that 2.5 cm.
Torque = 375 * 0.025 ≃ 9 Nm applied to your rear sprocket.
If you use a sprocket twice as big and apply the same force to the chain, you'll be applying twice the torque to the freewheel threads. However, if you want to go at the same speed at the same cadence with a bigger sprocket on the back, you're going to be using a bigger chainring. This will increase the velocity of the chain pulled. Remember that F=P/V; you've just doubled the V so halved the F. When we come to calculate torque = FD, we've doubled the D but since the F is halved, torque stays the same.