To make the new circuit work the same way as the existing one, you'd need to replace Q1 with a p-channel MOSFET, not an n-channel one.
However, you would not save any power dissipation by doing this. Because you wouldn't operate the MOSFET fully switched, you'd be operating it as a variable resistor.
You can see that there's no way to significantly improve the power dissipation of this circuit by the equation for device power: P = I * V. In this circuit, the current through Q1 is equal to the load current, so you can't change that. And the voltage across Q1 (collector to emitter) is equal to Vin - Vout, and you can't change that. So no matter what you replace Q1 with (however good a MOSFET you find), the power dissipation will still be Iload * (Vin - Vout).
If you want to reduce the power dissipation from your regulator, you need to look to a totally different type of circuit, like a buck switching regulator.