What he said ^^^ with one addition. The LED acts as a constant voltage source within certain current ranges, therefore you can reduce its brightness by decreasing the current flowing through it. the LED is probably drawing somewhere between 20 and 50 mA. This means that the inbuilt resistor is probably around the 240R to 600R mark. Resistors are worth a few cents each, so to half the current draw and reduce the brightness go buy say a 220R, 390R and 560R, maybe even an 820R, 1/4 or 1/2 watt resistor from Jaycar and try each one between the ground wire and ground until you get the brightness you like.