LED MOSFET switching circuit
The datasheet is your friend.

Assuming you are using an Arduino Uno with 5V outputs, when a GPIO pin is high, \$V_{GS} = \$ 5V. Your drain current \$I_D\$ is the same as the current through an LED, so you set this to your desired forward current. Let's say \$I_D = I_{LED} = \$ 5mA.

We now know that \$V_{GS}\$ = 5V and \$I_D = \$ 5mA. Figure 2 of the datasheet provides a nominal \$R_{DS(ON)}\$ at a given gate voltage and drain current. For \$V_{GS}\$ = 5V, \$I_D = \$ 5mA, the figure shows just a tad below 1.5Ω. Knowing this, we can approximate \$V_{DS} = R_{DS} * I_D \simeq \$ 7.5mV. This is small enough to neglect. Keep in mind that the datasheet values are nominal, and your MOSFET may vary quite a bit. Luckily, even if \$R_{DS}\$ varies by a factor of 10, it will still be negligible for your application!

In general, assuming you have a chosen an appropriate MOSFET (and are therefore operating with a \$V_{GS} &gt; V_{GS(th)}\$ and you have selected said MOSFET so that its nominal \$R_{DS(ON)}\$ is small (higher current implies a smaller needed \$R_{DS(ON)}\$), the \$V_{DS}\$ of the MOSFET when it is on should be negligible. In other words, you should select a MOSFET such that you don't need to factor it into your current-limiting resistor equation.


  Also, do I need any pull up resistor or anything to increase its efficiency or reliability?


Your Arduino GPIO pin will drive the gate both high and low, so no, a pullup/pulldown isn't necessary (as long as your Arduino will always be connected to the MOSFETs and driving them). However, it is good practice to put a gate resistor between the Arduino and each MOSFET gate. In this application, it's not crucial, but a gate resistor controls ringing at the gate. 100Ω or so is fine (the value isn't crucial).

If the Arduino isn't always connected, then a pulldown resistor is appropriate (to prevent the gates from floating). 1MΩ to ground is sufficient. Again, the value isn't particularly crucial, any large resistor is fine.