First, you need to know the electrical characteristics of your LEDs. For this, I will assume the forward voltage is 2.0V, and the forward drive current is 20mA. (.02A)
To wire LEDs in series, the cumulative forward voltage of the LEDs should not exceed the peak voltage of the power supply. I assume 14.4vdc as the peak supply voltage in an automobile. Exceeding the supply voltage will cause the LEDs to glow dimly. If the LEDs you are using are all the same, you can just multiply the number of LEDs by the forward voltage. (I recommend using the same type of LED in a series circuit)
To determine what resistor you will be needing, subtract the TOTAL forward voltage required to light the LEDs from the supply voltage. Divide this number by the forward drive current. The resulting number is the resistance value in Ohms of the resistor that is needed for that string of LEDs. (Be sure to pay attention to order if operations in the formula!)
Now, you will need to determine what wattage the resistor should be. Take the LED forward drive current and multiply it by the supply voltage. The resulting number is the power (in WATTS) that the circuit will draw. You should size your resistor so that it's rated wattage is higher than the power that the circuit will draw. Essentially, a 1/4watt resistor has enough mass to dissipate .250 watts worth of heat safely. If you try and push more power through it than it is designed to handle, the resistor will get very hot, and possibly cause a fire.
Since LEDs will only light if current flows in the right direction, and can easily be damaged if they are wired backwards, you need to pay attention to the orientation of your LEDs. TYPICALLY the flat-spotted side of the LED base is the (-) side, and the longer of the two leads is the (+) side. I have a few cheap Taiwanese LEDs around that are opposite, but they don't follow the rules...
So, to light a string of 4 LEDs off a 14.4c (peak) power supply:
14.4v - (4 x 2.0v) / .02A = 320
14.4v x .02a = .288 (So a 320-Ohm, 1/2 watt resistor needed)