Yes, a series resistor is one way you can do it. Measure the current when it is running at the desired brightness, in this case at 3.3V if you've settled on that, then plug the numbers into an LED calculator. You should also decide whether you want that target brightness at the vehicle (alternator) running state voltage, which you could assume is around 14.4V or measure it, then have it a little dimmer at the vehicle off state voltage of 12.6V (assuming 12V battery), or vice versa.
https://www.digikey.com/en/resource...ors/conversion-calculator-led-series-resistor
So if you want the brightness fixed at 14.4V the site above would have:
Supply Voltage 14.4
Forward Voltage 3.3
Forward Current (in mA) - whatever you measure with your 3.3V bench test, in this example I input 150 (mA).
Resistor Value 74 ohms (round up or down if needed to arrive at a popular value)
Power 1.665W - This is close enough to 2W, that I would move up to a 3W resistor instead of 2W to make the thermal density lower. It will get hot.
A buck current regulating switching PSU module would be more efficient, and are only $2-3 on ebay (with a month wait for shipping from China) but if talking about only a couple watts loss, then a resistor would be acceptable to me. I would consider where you put it since, again it will get hot.
Are you going to leave the resistor unmounted or mount it? Heatshrink tubing might not last long if that's the only thing you depend on to insulate it thermally and electrically from its surroundings, so if you have large diameter heatshrink tubing to fit the whole resistor inside (if you're not mounting it) you could just go with a higher wattage rated resistor, or you could split up (spread out over a larger area) the heat using multiple resistors in series
or parallel to reach the resistance value you need.
If you're mounting it, you could get a resistor that's heatsink-able and mount it to chassis metal somewhere which will heatsink it some and by its sheer size (and corresponding overkill wattage rating) will run cooler. The heatshrink tubing would then only cover the wire and terminals of the resistor, not the whole body of it. Here are some examples:
https://www.ohmite.com/heat-sinkable/
Remember that much of what I wrote above, depends on that 150 mA value I randomly choose to input into the Digikey calculator to result in 1.665W. Do the calculation using your own numbers to see how much resistor wattage you will have and go from there.
It is always good to add a fuse to vehicle add-on circuits, where you make the new tap into the existing electrical system rather than at the load so everything added has that protection. You may already know this, but a random site visitor might not think about it.
Edit: One other thing to keep in mind is that if you are using multiple glow strips, connecting them in series will mean a higher forward voltage and so a lower voltage drop from 14.4V, so you could use a lower ohm resistor that has lower wattage and heat dissipation. Just use the sum of their forward voltages which in the example I posed above would be (for two strips in series, 2 x 3.3V = ) 6.6V as the forward voltage in the LED calculator.
