Simple answer, use a permanant marker and blacken the LED.
If you can't see it, scratch some off.
Seriously, though. Just to clarify, and I don't mean to sound condescending, just letting you know, Voltage has potential. Current (amps) is flow.
LED's are diodes (Light Emitting Diode) and have a breakdown voltage of around 1.7V. In most 12V situations I think it is a 220ohm or 220K ohm resistor that is used for current limiting, so as not to blow the LED. If the voltage reaching the diode is less than 1.7V it will turn off. Theoretically a Diode is either on or off, but I have seen them dim and brighten depending on current.
You could try installing a potentiometer (or variable resistor for those not familiar with electronics) that will allow you to adjust the resistance. That way, during the day you can have it brighter and at night you can make it dimmer.
Or use the Pot to set a comfortable level, take it out, measure the resistance and buy a resistor in that area of value.
Or just buy a pack of various value resistors and try each one by one. Resistors are a dime a dozen these days.
Without extensive calculations, and measurements it can be difficult to pick a value and in this instance trial an error may yield quicker results.
Just an example on the calculations,
Calculating the Required Resistor
The calculation used to find the value of the series resistor we need to know the diode forward voltage and current and its connections. This information can be obtained from the package if you purchased the LED.
In this example it is 2 volts and 20mA (0.02 amps).
The cathode lead is the one nearest a "flat" on the body.
Since the voltage across the diode (LED) is 2 volts and the battery voltage is 12 volts, then the voltage across the resistor is 12-2 = 10 volts.
The diode is in series with the resistor, so the current through then both is the same, 0.02 amps.
We now know the voltage across, and the current through the resistor.
From Ohm's Law we can now calculate the value of the resistor.
Resistance = Volts divided by Amps = V/I = 10/0.02 =500 ohms.
Since this is not a standard value we can use a 470 or 560 ohm resistor as this application is not critical of values.
For a 4 volt (Blue or White LED) the formula would be:
Resistance = Volts divided by Amps = V/I = 8/0.02 =400 ohms.
The common 390 or 470 ohm resistor can be used for these.
Another thing, to avoid loosing too much voltage drop across the LED and causing it to turn off, is to put a resistor in Parallel to the LED, that way the voltage drop would be less affected by the value of resistor, mind you it does have an affect, but needless to say, with a resistor in parallel will sink some of the current, thereby reducing the current in the LED.
Calculating resistance in parallel uses fractions and I hate fractions!