Given a regulator with the same switching frequency, increasing or decreasing the output inductor value will do a few things.
First, Increasing the inductance will decrease inductor ripple current while decreasing the inductance will increase the ripple current:
Iripple = [(Vin-Vout)/L]*[(Vout/Vin)/fsw] ;where fsw=switching frequency
Modifying the ripple current will have a direct impact on the magnitude of the ripple voltage that is on the output rail:
Vripple = Iripple*ESR ;where ESR is the ESR of the output cap bank
Second, a larger inductor value will result in an increased amount of time for the regulator to slew up the current when a load is applied. When a load transient is applied to the output of a regulator, the output capacitors will supply the necessary current to the load. This results in a drop in the output voltage which the regulator then reacts to. The regulator must then supply enough current to satisfy the load and also to recharge the cap bank - the latter will only be a transient itself and the current requirement will back of a bit once the output is back up the the nominal regulation point.
A similar point can be made for load release. Once a load is released, the immediate result will be a slight increase in the output rail because the current through the inductor is no longer going to a load, so what it does is charge up the output capacitors. A smaller inductor allows for a quicker slew down of the inductor current which would result in a smaller overshoot of the output.
Finally, the output inductor is part of the output filter - the other part being the output capacitors. Significantly increasing or decreasing the inductance will have an effect on the response of the system which could require a retuning of the compensation network.