This is a nasty circuit, if you want to run high current LEDs from a high voltage. It will dissipate as much heat as a resistor sized to give the same current and voltage-drop, but managing the heat from this device is much harder.
So to calculate: multiple the current by the voltage drop to get the dissipated power. E.g. if you drive the blue LED segment at 350mA from a 12V supply, and the LED drops 3.2V at 350mA, then the power Pd = 0.35 x (12 - 3.2) = 3.08W.
Consider a buck-converter solution instead, which drops the voltage across an inductor without dissipating the sort of heat a linear solution does. For example, this part http://uk.farnell.com/diodes-inc/ap8801m8g-13/ic-led-driver-buck-0-5a-8msop/dp/1843875 is as simple to add as your TI part, allows PWM brightness modulation and just requires the addition of a suitable inductor. There are many more similar parts out there, try the parametric search offered by Farnell/Newark - I use it as a design tool daily in my work.