Voltages to keep my circuit running at 200mA with NMOSFETS
You do not need the extra resistors around M5, the circuit will work just as it is, but with R1 11.5 ohms. 

Vgs(th) for those MOSFET's is between 1V and 2V, so you just use 2V which is the worst case in this circuit. The RPI outputs 3.3V so it can easily turn on M5. When M5 turns on its drain will be almost 0V, which means the source of M1 to M4 will also be 0V. The RPI can easily turn on M1 to M4 because the 3.3V output from the GPIO exceeds the Vgs(th). When M5 is turned off then it doesn't matter what happens with M1 to M4 because no current can flow in any case.

In addition I suggest that you do not need M5 for the circuit to work. You can send the IR data to one MOSFET M1 to M4, while keeping the other MOSFET's turned off.

Answers to your questions

How do I calculate the resistors required for the gates, in order to lower the voltage to 2V? 

You don't need to lower the gate voltage to 2V. Unlike a BJT where you have to control Vbe, a MOSFET works with any gate voltage up to its maximum Vgs (this is for a digital circuit which is on or off). Just apply 3.3V from the GPIO. A MOSFET takes almost no gate current - it is specified as Gate-Source leakage current with a maximum of 10uA but is typically much less. 

You don't actually need a gate resistor, since you don't need to drop any voltage. However, it is still a good idea to add one, especially to larger MOSFET's. The gate of a MOSFET looks like a small value capacitor which has to be charged up to turn the MOSFET on, which can result in a large current flow from the GPIO pin until the cap has charged. When I am driving it from a processor I prefer to limit this current to be within the current rating of the GPIO pin. In this circuit I would limit it to 10mA with a 330 Ohm resistor. 

Also, Why is M5's drain almost and not exactly 0V? 

When it is turned on the MOSFET has a small resistance between drain and source (rDS(ON) ID) which will drop a very small voltage when current flows through it. Hence, its drain is nearly 0V. In this circuit you can ignore it and assume it is 0V.