Steve,
So I have learned a lot in this exercise, which is really why I enjoy doing this anyway. BB, I can appreciate your sentiment, but I am not doing this for the final number; rather, I just cant resist a fun problem. For me, figuring things out just gives me enjoyment; it's probably wierd, but that's just me. Of course, in practice, if I was really interested in a fan's power consumption, I'd just put a wattmeter on it and measure it. But, doing the physics is just plain entertaining!
After re-checking the calculations, it looks like this is indeed the correct equation for the minimum power you need to move a given quantity of air. The density of air is a variable, but it doesn't change things much. But I have learned a lot about fan efficiency in general.
So, we can run the numbers. First point, 20 W is inconsistent with 220mA at 120V, since the latter would give 26.4 W.
As far as area, the intake is really not square, it is round. The fan blade diameter is 4 inches, since I think your numbers are for the housing size.
For a 2 inch radius, we have a area of (3.14)*(2*2) = 12.6 square inches. Now, we must subtract the hub, since no air moves from there. I am guessing the hub radius is 1 inch. That gives a hub area of 3.14 square inches, for a net intake area of 9.46 square inches.
How much power would we need to move air at the rate of 65 CFM through a opening of 9.46 square inches if we could move it with 100% efficiency? That is the question the equation will answer. So we plug in the numbers -
P = (0.0001565)*(65*65*65)/(9.46)*(9.46) = 0.48 W.
Now, this is a long way from the 20W rating the fan claims. And I agree. There are two factors here.
One, the fan is not 100% efficient. This is really where I learned something. Fan motors, with their current designs, aren't very efficient. Most of the energy goes to heat, and 12% to 15% actually is transferred to the shaft! So if we assume a 100 percent blade efficiency, the overall fan efficiency is still 15 percent. So to move 65CFM using this fan, we would need a power of 0.48W / 0.15 = 3.2 W.
Still a long way from the 20W rating - well yes. When the fan starts, the blade has to get moving, which needs energy. You can figure this out too! But it is a bit complicated, as you need to compute the moment of inertia of the blade, and know the fan's steady state rpm. We'll leave that calculation alone, but I'm willing to bet that is where the extra 17 W or so goes. The published rating has to handle the worst case, which includes the extra juice needed for fan startup.
I'd say if you measured the current being drawn by your fan while it is running, it will be a whole lot less than 220mA. It will be closer to 22 mA. I'd try it out and see.
What is the point in all of this - well, I'll spin it this way - we don't think too much about improving fan efficiency because the power consumption is low anyways. But given it is 15 around percent max, we are being quite wasteful. We can go a long way with better motor and blade designs.
From a scientific standpoint, it is just fascinating to learn that, other things being equal, if you need to get twice the CFM without making the fan blades bigger, you will need a motor that sucks eight times the juice! Or that if you use a fan intake twice as large in area, you will only need a quarter of the energy to keep the same CFM. Someting I didn't know. Before this, I'd have thought nothing of reducing the area from which air can go in through a fan housing, but as you can tell, for a given fan motor, if you reduce your intake area by just 20 percent, the CFM will drop by 34 percent!
I'll post how I came up with my numbers in the ash can when I get a chance.
I got my hopes up. 
