Electrical – Proper resistor(s) for limiting LED brightness

Without a schematic, based on your question, you have Common Cathode Led with TIP120 transistors? I assume you have them connected backwards from normal.

^{simulate this circuit – Schematic created using CircuitLab}

The TIP120 is a NPN darlington pair transistor. It normally expects to be on the low side of the load. You are using them for High Side Switching, which won't work. If your led was common anode, you could swap them around, and it would work (WITH THE RESISTORS).

But in this case, you have 1) A different signal level from the led power level (3.3v vs 5v) so Direct PNP transistor won't work, and 2) The Raspberry PI GPIO cannot source a lot of current. 16mA at max. So You need both NPN and PNP transistors. Some 2n3904 and 2n3906 Common npn and pnp transistors, but any similar would work

^{simulate this circuit}

I only show the blueled (5v source voltage - 3.2v forward voltage - 0.2v VCE drop / 20mA = 80Ω), do the same for the green led, and use a 140Ω or higher resistor for the red led.

Yes, LED strips have resistors built in.

Resistors are primarily rated by ohms (Ω), and secondarily by power dissipation. Standard component resistors are rated at 1/4W dissipation. That just means that you can't put more power through them without burning them out.

To figure out what to put in the resistor block of your circuit we'll just use Ohm's law (V = I R):

1. You should measure with an ammeter to be sure, but 5050 LED strips are typically assembled with resistors to draw 1 amp per meter at 12V, which means that the strips have built-in resistance of 12Ω.
2. Decide how much "dimming" you want. If you want them at quarter brightness, and noting that luminosity is proportional to current, then quadruple the resistance by adding a 36Ω resistor, which will bring total resistance to 48Ω and reduce current to .25A/m.
3. But you can't just pop any 36Ω resistor in front of the load. Suppose you have 1 meter of strips on your helmet: That would mean that you have to dissipate .25A*12V ≈ 3W of power. Therefore, either find a ≥ 36 Ω 3W resistor, or else use twelve standard 3 Ω 1/4W resistors in series, or twelve standard 432 Ω 1/4W resistors in parallel, in order to dissipate the power used by a meter of those LEDs.