I understand you want to be able to simultaneously feed 20 mA through each subpart of 120 RGB LEDs. That is a total of 7.2A at (maximum) 14.4V, for a grand total of 104W, of which 24W is dissipated in the LEDs. If you exclude switching voltage conversion the remaining 80W of heat must be dissipated somewhere.
A first step is to establish the maximum ambient temperature your chips must tolerate without going into thermal protection mode. Then you can use fig 11 in the datasheet to find the power a single chip can handle. Let's for the moment assume 2W. You need 120 * 3 / 24 = 15 chips, so they can handle 30W. That leaves 50 W to dissipate elsewhere.
You could insert a resistor in series with each LED. Assuming a worst case drop over the LED of 4V, 1V for the chip, and a worst case low battery of 10V leaves 5V for the resistor, which must hence be 5 / 0.02 = 250 Ohm. Calculating from the opposite case, 2.4V for the LED, 14.4V accu, this gives 14.4 - 2.4 - 5 = 7V for the chip, so it dissipates 7 * 0.02 * 24 = 3.4W. That's still uncomfortably high.
A better solution would be to use a regulated (switched!) 5V supply. Now the chip has to dissipate (assuming a worst case LED drop of 2.4V) 2.6V, for a total dissipation of 2.6 * 0.02 * 24 = 1.25 W. That's more comfortable. And you don't need the resistors (but in exchange for a single SPSU).
My calculations show how you can evaluate these two designs. It is up to you to supply the correct figures (ambient temperature, range of LED drop, accu voltage range, etc), redo the calculations, and evaluate the results.