Why do I get such high values when calculating the power dissipation of MOSFET in boost-converter circuit?
I wonder if it's possible to calculate how much the mosfet is going to
heat up

14.93 amps is the correct figure for the DCM peak inductor current but then your calculations go wrong. You cannot calculate the resistance of the inductor that way but, it's pointless anyway - you are trying to find the MOSFET power dissipation and so the resistance of the inductor is irrelevant.

Use the red waveform above and calculate the power dissipated in the MOSFET's conduction period (\$D\cdot T_{SW}\$). Then average that power over one full cycle (\$T_{SW}\$). That's how much heat will be generated by the MOSFET.
What you will find is that if the MOSFET waveform were a continuous saw-tooth waveform then the RMS current is: -
$$\dfrac{I_P}{\sqrt3}$$
And the power would therefore be: -
$$\dfrac{I_P^2\cdot R_{DS(ON)}}{3} = 19.32 \text{ watts}$$
But this power is only dissipated for 78.38% of the time hence the real power dissipated by the MOSFET is 15.14 watts.
I assumed 0.26 ohms for \$R_{DS(ON)}\$ in the above calculation because the gate drive voltage might be 12 volts instead of the 10 volt figure in the data sheet. I would also highly recommend that a device with significantly smaller on-resistance be used because the IRFP460 is pretty shabby in this respect.

Calculating RMS of a voltage or current triangle or saw-tooth waveform.