Calculate 12v Battery Runtime When Using Only 3.6v
Ok, we have:


12 LEDS with Vf=3.6V and Iftot=320mA when in parallel (If=24mA per LED).
12V battery with let's say 1000mAh (you haven't specified the capacity)
The battery voltage will be regulated down to 3.6V with voltage regulator


Assuming you use a switching regulator with 100% efficiency we can calculate:

Ibat = Vf / Vbat * Iftot = 96mA

Thus the battery would last:

t = 1000mAh / 96mA = 10.41h

Since a switching regulator has more like a 80-90% efficiency, expect 10-20% less.

If you use a linear regulator like the LM-series, the difference between the input and regulated voltage is dissipated as waste heat. Then it would just be:

t = 1000mAh / 320mA = 3.12h

And your regulator would get pretty hot (P= 12V - 3,6V / 320mA = 2.7W).