There are many losses associated with switching, but it sounds like you are most concerned about the additional thermal load introduced into the MOSFETs in the period transitioning between on and off. I thought it would be easy to find some application notes on this, but surprisingly it wasn't. The best I found was AN-6005 Synchronous buck MOSFET loss calculations with Excel model from Fairchild, the relevant parts of which I'll summarize here.
During the switching transition, the voltage and current in the MOSFET will look approximately like this:
The switching losses we are going to calculate are those in periods \$t2\$ and \$t3\$ due to the voltage and current in the MOSFET. The way to approach this is to calculate the energy of each transition, then convert this into an average power according to your switching frequency.
If you look at just \$t2\$, \$V\$ is nearly constant, and \$I\$ increases approximately linearly, forming a triangle. Thus, the power also increases linearly, and the total energy is the time integral of power. So the energy is just the area of that triangle:
$$ E_{t2} = t_2 \left( \frac{V_{in} I_{out}}{2} \right) $$
\$t3\$ also forms a triangle. In this case, the voltage is changing instead of the current, but still the power makes a triangle, and the calculation of energy is the same.
Since the calculation is the same for \$t2\$ and \$t3\$, then it's not really important how much time is spent in \$t2\$ vs \$t3\$; all that really matters is the total time spent switching. The energy losses from one switch are thus:
$$ E_{switch} = (t_2 + t_3) \left( \frac{V_{in} I_{out}}{2} \right) $$
And, your switching frequency is how many times per second you incur this energy loss, so multiplying the two together gets you the average power loss due to switching:
$$ P_{switch} = f (t_2 + t_3) \left( \frac{V_{in} I_{out}}{2} \right) $$
So, taking your calculation of the switching period being \$150ns\$, and the maximum current being \$330A\$, and the voltage \$12V\$, and the switching frequency \$30kHz\$, the power losses from switching are:
$$ 30kHz \cdot 150ns \left( \frac{12V \cdot 330A}{2} \right) = 8.91W $$
That's \$8.91W\$, ideally, shared between three transistors, so only about \$3W\$ each, which is pretty insignificant compared to your other losses.
This number can be checked for sanity with a simpler model: if you spent \$150ns\$ switching, and you do it \$30000\$ times per second, then you can calculate the fraction of the time you spend switching, and make the most pessimistic assumption of the full power of \$12V\cdot330A\$ being lost in the transistors:
$$ \require{cancel}
\frac{150 \cdot 10^{-9} \cancel{s}}{\cancel{switch}}
\frac{30 \cdot 10^3 \cancel{switches}}{\cancel{s}}
\cdot 12V
\cdot 330A = 17.82W $$
Of course, over the switching period, the average current and voltage is only half that of the maximum, so the switching losses are half this, which is what we just calculated.
However, I bet in practice, your switching times will be slower. A "\$2A\$ gate driver" isn't a constant current source as these calculations assume. The real picture is rather more complicated than this simple model. Additionally, the current will be limited by the resistance, and usually more significantly, the inductance of the transistor packages and the traces leading to them.
Let's just say the inductance of the gate driver, transistor package, and traces to it is \$1\mu H\$. If your gate drive voltage is \$12V\$, then \$di/dt\$ is limited to \$12V/1\mu H = (1.2\cdot 10^7)A/s\$. This may seem like a lot, but on the time scale of \$150ns\$, it's not. Keeping the inductance low will take some very careful layout.
So, I would say that these calculations show that your switching losses may be manageable, though you won't know for sure until you've made the layout and tested it. Even if you can't reach the ideal of a \$150ns\$ switching time, the losses are low enough relative to your other problems that you have some margin to do worse and still function.
Your bigger problem is probably getting the three MOSFETs to switch at the same time. Otherwise, one of them will get a disproportionate share of the total current, and thus heat, leading to premature failure.
Best Answer
Driving with TTL is a bad idea. At 4D drive the IRFZ44 is only guaranteeing to pass 250 micro amps. You need to follow the guidelines in the spec: -
The spec says it has a test to check rise times and fall times - they use a 10V pulse with an output impedance of 12 ohms.
You cannot expect to get anywhere near this performance from TTL at 4 or 5V. The gate input capacitance is 1.5nF and this needs something like a 1 or 2A drive (off the top of my head) to get the device to switch on and off at the rate you are likely wanting.
EDIT to include drive current into the gate.
It's easiest to start off with Q = CV then differentiating we get \$\dfrac{dQ}{dT} = C\dfrac{dV}{dT}\$ where
\$\dfrac{dQ}{dt}\$ equals charging current into the gate capacitor of 1.5nF.
The voltage on the gate needs to change about 10V in 20 ns hence \$\dfrac{dV}{dt}\$ = 500,000,000.
Therefore charging current (to be supplied by gate drive) is 1.5 \$\times 10^{-9} \times 500,000,000 = 0.75A\$. This means your driver ought to be able to deliver 1 or 2A as previously mentioned.