You need a proper value resistor. For example, for a 2.2 volt LED, you want a resistor of
12 - 2.2
------------- = 490 ohms
That's from Ohm's law, R = E/I, where R is in ohms, E is in volts, I is current in amps.
470 ohms would be a close standard value. This would give a current of not quite 21 mA, which is fine.
HOWEVER. While the car is operating the voltage at the battery terminals can be as high as 14.4 volts, so that's the value you should use in the calculation:
14.4 - 2.2
-------------- = 610 ohms
620 ohms is a very close standard value. Power dissipation is now 0.24 watts so I'd strongly recommend a half watt resistor. (You always want a safety margin.)
If you are running this while the car is turned off, the battery voltage may be close to 12 volts. In that case your current will be less. How much less?
12 - 2.2
----------- = .0158 A = 15.8 mA
...a bit less than you'd like, but you are unlikely to notice the brightness difference.
Once you have the proper resistor value in the circuit, current beyond what the resistor allows simply cannot flow. It does not matter how much current the battery can supply. Consider that it is perfectly fine to plug a night light that consumes .06 amps or so into a wall outlet that can supply 15 or 20 amps. It's a similar situation here. The LED is a "strange" component - it behaves in a circuit basically like any other diode - but as long as the resistor is properly calculated the current will be limited to I amps, where I = E/R.