Is there a limit to capacitor peak power density?
Your post is unclear so I just offer the following observations.


  Wikipedia's Power to weight ratio
  mentions a 3300 V, 100 kA capacitor, which works out to 330 MW.


That is in the section Electrostatic, electrolytic and electrochemical capacitors and the details are:

Type                      General Atomics 3330CMX2205 High Voltage Capacitor
Capacity                  20.5 mF
Voltage                   3300 V
Energy to weight ration   2.3 kJ/kg
Power to weight ratio     6.8 MW/kg @ 100 kA


330 MW is not "the power of the capacitor". It is the peak power the capacitor can deliver. 

We can work out roughly how long this will last as follows:

$$ Q = CV $$
$$ V = \frac {Q}{C} $$
$$ \frac {dV}{dt} = \frac {dQ}{dt}\frac {1}{C} = \frac {I}{C} = \frac {100k}{20.5m} = 4.8 \ \text {MV/s} $$

Your capacitor, if it discharged linearly would be flat in \$ \frac {3300}{4.8M} = 0.7 \ \text {ms} \$.

However, I suspect that 100 kA can only be delivered into a short circuit and in that case the output voltage is zero so the power delivered to the load is zero and all the energy is dissipated in the internal resistance of the capacitor. This is probably the main flaw with your thinking. If at 3300 V the device can supply 100 kA max then it implies the internal resistance is \$ \frac {V}{I} = \frac {3300}{100k} = 33 \ \text m \Omega \$.

Now, using your maximum power transfer idea we would use a 33 m&Omega; load, current would be reduced to 50 kA, voltage reduced to half of 3300 and the maximum instantaneous power to the load would be \$ 50k \times 3.3k /2 = 82.5 \ \text {MW} \$.