Method
I'm using IndustrialCraft v5.12 (MC beta 1.4_01).
I cut a 5x5x5 tank out underground.
I put a water mill in the center.
I used 4 wires directly beneath the water mill.
I dug an access shaft beside and under the tank.
I will be placing and removing the MFE from the side of the 4th wire.
F3 didn't report any other frame rate besides 58-60 fps.
I'm not going to use IC's "s" unit so "sec" means an actual second.
5x5x5 - 3
I begin timing as soon as I connect the MFE.
5 minutes (4:59) after placing the MFE read 2020.
I removed the MFE.
3x3x3 - 2
I then built over the tank walls and lowered the top layer of water to give a 3x3x3 cube.
I begin timing as soon as I connect the MFE.
5 minutes (5:02) after placing the MFE read 2108.
I removed the MFE.
3x3x2 - 1
I then built on the tank floor to give a 3x3x2 cube.
5 minutes (5:00) after placing the MFE read 1412.
I removed the MFE.
Errata
I accidentally ran the 3x3x3 trial while the tank was empty and still got 8 EU in the MFE after 5 minutes.
Experimental Data (time, EU, water volume)
t1 = 239 sec
q1 = 2020 EU
v1 = 122 cubic meters
t2 = 302 sec
q2 = 2108 EU
v2 = 25 cubic meters
t3 = 300 sec
q3 = 1412 EU
v3= 17 cubic meters
Calculated EU / sec rates:
r1 = 8.45 EU/sec = 2020 EU / 239 sec
r2 = 6.98 EU/sec = 2108 EU / 302 sec
r3 = 4.71 EU/sec = 1412 EU / 300 sec
Calculated rate density
d1 = 0.070 EU/sec / cu m = 8.45 EU/sec / 122 cu m
d2 = 0.279 EU/sec / cu m = 6.98 EU/sec / 25.0 cu m
d3 = 0.277 EU/sec / cu m = 4.71 EU/sec / 17.0 cu m
Edit: wrote t1, t2 but meant r1, r1; fixed.
The fact that r1 and r2 are almost the same suggests that the same amount of EU is stored for a 5x5x5 water cube as is stored for a 3x3x3 and that supports the hypothesis that only the 3x3x3 cube is contributing the the water mill.
The fact that d2 and d3 are almost the same suggests that each water source block contributes a constant rate of EU. Since we have already established the likelihood that only the 3x3x3 cube is counted, we can find more support for both hypotheses if we recalculate d1* to account for the smaller volume and find that it is also around 0.3 EU/sec / cu m and this calculation shows very nearly just that:
d1* = 0.338 EU/sec / cu m = 8.45 EU/sec / 25.0 cu m.
I don't just have my paltry data. Okamiltan built a 9x5x3 area and scattered 8 water mills. Okamiltan didn't tell us about his wiring but I'm going to assume he has 1 wire under each water mill:
http://www.minecraftforum.net/viewtopic.php?p=3726432#p3726432t4 = 350 sec
q4 = 10,000 EU
v4 = 119 cubic meters
r4 = 3.57 EU/sec = (10,000 EU / 350 sec) / 8
Firstly, r4 is much lower than r2:
51.1% = r4 / r2.
That alone is enough to suggest that tightly packing water mills diminishes their purpose.
That gave me hope that each water source block can only be used by one water mill and if that were the case then the rate density for Okamiltan's trial should be around 0.3 and but it was more than 200% more:
d4* = 0.706 = (10,000 EU / 350 sec) / 119 cu m
So it is a little more complicated than that. The rate density for each of Okamiltan's individual tightly packed water mills:
d4 = 0.143 EU/sec / cu m = 3.57 EU/sec / 25.0 cu m
So each individual mill gets about half as much EU/sec out of its 3x3x3 cube but the overlapping of these mills means you can actually double the EU/s that you can get out of each block of water but this only makes sense if water source blocks were much more expensive than water mills.
I'm afraid I don't have the time to really work out in more detail what is actually happening in suboptimal patterns so I will just leave that as an open question.