LED Driver IC off a single-cell Li-ion battery (Too large of a voltage difference?)
Your maths is not quite correct. You are right that the LED will be drawing 4.9W, and most of this will be converted to heat (about 80%). You are also correct that the current draw from the battery to power the light will be 1.6A. However, you do not multiply the voltage difference to get the heat dissipated by the IC. This is a switching converter, so the heat dissipation to the chip is a function of the efficiency of the chip, which depends on the mosfet resistance, switching speed, and other things (plus the inductor ESR). Usually the datasheet will give you an idea of the efficiency at different currents and voltages. Looking at the graphs, its probably around 80%, so your heat in the chip will be about 1.2W, and your total current draw from the battery will be 2A.
You are also correct though that this chip will not be able to drive 700mA, since you need 2A inductor currents.