Can N-Channel MOSFETs safely discharge capacitors?
Weather they can handle the discharge depends on a few things.
1)  How much energy is in the capacitors. E = 0.5 * C * V^2

2)  Where do you expect the energy will be dissipated during the short.  And what fraction of the total energy is going into the FET.  Energy may be dissipated in the following places.

  a)  In the internal ESR of the capacitors.
  b)  In the drain source resistance of the FET
  c)  In an external series resistor put in the circuit to absorb energy during the short
  d)  In the wiring resistance

3)  What pulsed energy can the FET handle.  You can usually find this in the Safe Operating Area graph for the FET.

For example...
C = 1mF 
V = 30V
ESR = 15mOhms (from capacitor datasheet)
Therefore...
E = 0.5 * 1mF * 30V * 30V = 450mJ

Lets say that the datasheet for the FET lists Rds_on = 10mOhms.
Lets also say that we have 5mOhms of wiring resistance.

The peak curent would then be 30V / (10mOhms + 5mOhms + 15mOhms) = 1000A.

Lets say that the FET datasheet safe operating area graph says that the FET can handle 1000A with Rds = 30V for 10us.  The pulsed energy rating would then be 0.3J.  If the graph didn't say it could handle a 1000A pulse we would need to add a series resistor.

The energy dissipated in the FET is 450mJ * 10mOhms / (10mOhms + 5mOhms + 15mOhms) = 150mJ.

So in this case the FET would survive because 150mJ is less than 0.3J.

Lets try another example...
C = 1mF 
V = 50V
ESR = 15mOhms (from capacitor datasheet)
Therefore...
E = 0.5 * 1mF * 30V * 30V = 1.25J

Lets say that the datasheet for the FET lists Rds_on = 10mOhms.
Lets also say that we have 5mOhms of wiring resistance.

The peak curent would then be 50V / (10mOhms + 5mOhms + 15mOhms) = 1666A.

Lets suppose that the FET datasheet says the FET can't handle more than 100A peak.  Therefore we add a 0.5 ohm resistor in series to limit the current.

The peak current is now 50V / (10mOhms + 5mOhms + 15mOhms + 0.5 ohms) = 94.34A.

Lets also say that the SOA graph says the FET can handle 100A at 50V for 100us.  100A * 50V * 100us = 0.5J.

The energy dissipated by the FET during the discharge is...

1.25J * 10mOhms / (10mOhms + 5mOhms + 15mOhms + 0.5 ohms) = 23mJ.

In this case adding the series resistor greatly reduced the energy dissipated by the FET.  Since 23mJ is much less than the 0.5J rating the FET in this example would survive.

One thing to point out in your specific case is that you can't depend on the Rds_on of the FETs being matched.  That is you can't assume that each FET will share current evenly.  Instead of 100A, 100A, 100A, you could get 70A, 130A, 100A or similar.  You can only guarantee that the current sharing is even if you add some external resistance that is significantly higher than the FETs Rds_on.