How Do I Calculate The Power Of A Rail Gun?
Based on Brian Drummond's comment: 


  A quick search will find a simple formula relating capacitance, voltage, and energy. To get from energy to power, you need to know how fast you are charging and discharging the caps. Don't forget that connecting 12 2200uF caps in series will give you about 180 uF, a rather dangerous voltage, and no more energy storage than connecting them in parallel.  


The maximum energy your capacitors will be able to store is:
E = 1/2 C V^2
Where C is the capacitance, V is the voltage the capacitors are charged to, and E is the energy stored. Using this, if all of the capacitors are charged to their maximum voltage, you will have an energy of:
(1/2)*2200*12*400^2 = 0.0011 * 12 * 160000 =.0132 * 160000
= 2112 Joules
Assuming an approximate weight of .989 oz, or mass of about 28g, and perfect energy transfer(impossible), the velocity necessary to have the same amount of kinetic energy as stored electrical potential is:
E_k = 1/2 m v^2
where E_k is kinetic energy, m is mass, and v is the velocity. so
2112 = (1/2)(.028kg)v^2
150857 = v^2
= 388 m/s or 867 mph.  

Now, given the highly imperfect nature of home building your own rail gun, I would say you need to be very lucky to get even half of that energy transferred to your nail, the case would instead be
1050 = (1/2)(.028kg)v^2
75000 = v^2
= 273 m/s or 610 mph.  

Even that figure is very optimistic, and ignores all energy lost to friction, air resistance (which is no where near negligible at these speeds), and even energy that wasn't able to get transferred to the nail before it leaves the barrel.  

One more note on Brian Drummond's comment:
The capacitors will store the same amount of energy in series or in parallel (assuming the voltage across every individual capacitor is kept the same), however, current will rise to a more useful level much faster in the series configuration (since the coil of the gun is just a long inductor) which means more energy has the opportunity to be transferred before the nail leaves the barrel.
These are very rough estimates that assume perfect conditions. Better estimate could be gotten if we knew the exact inductance of your coil, the exact mass of the nails you plan on firing, etc. But that would require some calculus that I don't want to do.    

Be safe, I don't want to hear about anyone getting killed by their own rail gun on the news.
Also, its possible my math is off (please correct me if it is), its been a while since I've really done classical physics.

EDIT:
I hadn't done much research on the subject, I simply used the formulas I knew to get approximations. After looking into the subject a little more you probably mean a coil gun instead of a rail gun. They use the same principles to work, but the construction of a rail gun is much more complex, requiring the use of a sliding armature and more advanced mechanics. A coil gun is simply an object accelerated through a tube by a magnetic field, from a coil around the tube.

On efficiency http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA389605 this document discusses the efficiency of rail guns, (lets ignore the fact that it is questioning if the efficiency is a useful metric), and states that typical efficiency is around 40% (100% minus 5% residual inductive, 20% lost in the armature, and 35% lost to friction [what they dubbed "rail resistive"]) I would hazard a guess that your best efficiency before any air resistance or other unknowns will be somewhere from 20-30%. Not great efficiency, but still enough to get a nail moving pretty fast.