The strips use 3V white LEDs, power to which is delivered via resistors or linear current regulators. Unless you see any inductors, there is no buck converter from 5 V to 3 V.
Why does this matter? Well, with resistors and linear regulators, 𝐼in (current in) pretty much equals 𝐼out (current out). So the efficiency 𝜂 = 𝑃in / 𝑃out = (𝐼 × 3 V) / (𝐼 × 5 V) = 60 %. Extra cable resistance will reduce the current, brightness and power, but still exactly 40% of power leaving the USB charger will be wasted before it gets to the LEDs.
However, I would advise against multiple, cheap USB connectors in the circuit: when moving the setup, their resistance changes somewhat and you would get blinking. The worst thing that could happen is a switch, such low voltage cannot spark over an oxide layer and eventually even small movement will blink the lights. I would get a good thick USB cable and solder it directly to one of the strips instead of whatever it came with, connecting the other with some thick wires.
So it does not matter, if you want better efficiency, use 12V (75%) or 24V strips (87%), or get just an LED array without resistors that needs a constant current driver (theoretically 100% but CC PSUs are slightly less efficient). Or make a constant current driver by fine-tuning the voltage of a PSU (by adjusting the feedback resistive divider) to 0.5-1 V above the LEDs’ voltage drop, then using an appropriate resistor to limit the current.