Inductor Inrush Current Rating
Yours is a reasonable calculation.

The principal effect of saturation is that your inductor temporarily changes inductance value from big to small. This is not an issue during the inrush period. As the magnetic material used in inductors is soft, it has a narrow hysteresis loop, it will recover and start working properly when the current drops back to spec.

Taking the inductor to a high magnetic field and then back again will dissipate a little energy in the core through hysteresis, but as you are doing it only once, and as we've said above the loop is very narrow, it's likely that the energy deposited will be negligible, probably much smaller than your one time \$I^2R\$ heating in the inductor resistance.

Bulk resistors, like wirewound resistors, and inductors, are very robust to pulse loading. The heat is deposited uniformly throughout the mass of wire. Components like diodes are much more fragile, where it's possible for all the heat to be dumped in a tiny junction area. As such, as long as the total energy deposited is limited, which keeps the temperature rise down, the power for wirewound components can be exceeded for a moment with impunity.

It's easy to measure the copper temperature of wound components. The tempco of copper is about 0.4%, or a 10% rise in resistance for every 25C rise. It might surprise you to do some experiments on your inductor, and calculate its \$I^2t\$ rating (it's unlikely to be given in the specs). Apply higher current pulses to it until its measured temperature rise for a single pulse gets to (say) 50C (very safe) or 75C (still pretty safe). It's important to use a short high current pulse, so the inductor heats without (to first order) losing any heat to the environment. Now compare that to the sort of energy you're considering, and understand why the data sheet doesn't bother to tell you that.