As I said before, resistors do not reduce voltage. They drop voltage. They drop the difference of whatever voltages are present on either side of them.
Diodes are different and will only drop whatever their forward voltage is. Most diodes are around 0.7V. (Light Emitting Diodes, LEDs, can be between 2 and 4V depending on the color.) Regardless of what voltage you apply to the diode, it will only drop its forward voltage. They will conduct whatever current is available, which is why you need a current limiting resistor in series with them (depending on the application.)
This is important to understand because...
My goal is to limit, as much as possible, the current used by each pin on my Arduino by offloading each circuit's output device with a darlington.
A darlington is a pair of BJT transistors. In terms of voltage drops, you should consider each BJT's Base to Emitter a diode's worth of voltage drop, which is 0.7V. So for a Darlington pair (DP), you are around 1.4V.
That means if you put a resistor between the pin of your arduino and the input of your DP, the DP will drop 1.4V leaving a 5V-1.4V=3.6V to drop across the resistor. Using Ohm's Law, you can see that 3.6V/2.2Kohms = 1.63mA. Since the resistor is in series with the pin an the DP, only 1.63mA will flow.
BJTs are not really voltage devices. They are current devices. They amplify the amount of current flowing through them. The more current flowing from base to emitter, means more current is allowed to flow from collector to emitter.