will my linear regulator overheat?
1 amp from a 5 volt linear regulator fed with a supply of 14.8 volts quite simply means the power dissipated by the device will be 1 x 9.8 volts = 9.8 watts. Power dissipation in the device is defined by the volt drop across it (9.8 volts) and the 1 amp flowing through it.

I'd definitely consider a switching regulator - power efficiency is going to be about 90% so, with 5 watts out, there will be about 5.5 watts in and, from a 14.8 volt supply this means it will take a current of about 375 mA i.e. a lot smaller drain on your battery.