Geometric Interpretation: Parallel forms are harmonic
If $\nabla\omega=0$ then also $d\omega=0$, since $d\omega$ is the skew-symmetrization of $\nabla\omega$ (this is true for any torsion-free connection, whether it's Levi-Civita or not). On the other hand (just for Levi-Civita), if $\omega$ is parallel then also $*\omega$ is parallel (since $*$ is defined in terms of the metric). Thus also $d*\omega=0$, so also $\delta\omega=0$ and $\Delta\omega=(d\delta+\delta d)\omega=0$.