"The potential energy of a falling body is calculated by integrating the force across the distance. But the force is not constant as the altitude increases, it decreases as the inverse square of the product of mass and acceleration. The gravitational acceleration is not the same at geostationary orbit altitudes as it is at the earth's surface."

This is true, but there are a lot of details I didn't include- and certainly a lot of technical problems to overcome.
My stated amount of power was almost twice 2030 projected global electrical need. Because of the decreasing the gravity it could reduce that number by say about 1/4, so it's off the total predicted global demand of electricity in 2030 by around 1/2, so you need to double the volume if you want to make up the difference. So that's somewhere around 4-5' pipe instead of a meter pipe [and that's assuming a meter diameter pipe does 10 cubic meters per second- which just SWAG]. More significantly it doubles the volume used, so instead of 1000 to 10,000 year range [because one concerned about sea level rise of a couple feet] it's a 500 to 5000 year period.
Of course I'm not sure how accurate the formula:
POWER (kW) = 5.9 x FLOW x HEAD
Actually is. It depends upon how efficient hydropower generators are generally made today. I suppose I could look that up somewhere.
One also would have drag losses, one has to have pipes strong enough to handle the psi load. And you can't simply drop water from GEO to earth surface and expect to be able to generate power from all the energy at the Earth's surface. One would need a series of generator along it's entire length which slowed down the speed of the water so it was alway somewhere below supersonic speed [because at such high speeds, water will simply cut through anything].

Anyway, you are correct. I had thought the lessened gravity would about 1/2 but I think now it's closer to a 1/4.

Darn, now I have figure this out more.
So, at 6378 kilometers gravity is quartered, at 12756 it's 1/9th, at 19134 km it's a 1/16, and at 25512 km it's 1/25th, and 31,890 km it's 1/36th, and at 38,268 it's 1/49th. See:
http://www.astronomynotes.com/gravappl/s5.htm

So, at 6 Earth radius out something on Earth that weighs 4900 lbs weighs 100 lb. At 5 radius it weighs 136 lbs and then 196, 306, 544, and 1225 lb at 1 radius.

So, it you had a stick that was 6378 kilometers long and weighed 1 billion tons, how much would weigh if it was vertical? Well, it's going to weigh at least 1/2 billion tons.
The same stick at 6378 kilometers high will weigh 1/4 billion tons and if you put it vertical it will at least weigh 1/8th of billion tons and at 12,756 km the stick which weighs 1/9th a billion tons if vertical will weigh at least 1/18th of a billion tons.
So much fun. But how much more than half? I don't think it's more than 2/3. So .666 + .1665 + .074 + change. So the effective height would be 6,378 km times .91 which is
5804 km.
So, POWER (kW) = 5.9 x FLOW x HEAD is:

5.9 x 10 X 5,804,000 meters is 342,436,000 kW
or 342.4 giga watt
So, it's only going to provide about 1/10 global electrical power needs of 2030. Increase the flow by factor of 10 if you need this much power and worry about rising sea level in about 1000 years.