Let $(M,g)$ be a Riemannian manifold.
The canonical volume form $\mu=\sqrt{\det g_{ij}}\,\mathrm{d}x^1\wedge\dots\wedge\mathrm{d}x^m$ is parallel with respect to the induced Levi-Civita conection $\nabla$ on $\Omega^m(M)$ hence $\mu$ is also harmonic.
Using local normal coordinates I am able to see that $\nabla\mu=0$ and also, since $\delta\mu=(-1)^{m(m+1)+1}\star\mathrm{d}\star\mu=(-1)^{m(m+1)+1}\star\mathrm{d}(1)=0$ and $\mathrm{d}\mu=0$, that $\Delta_{\operatorname{Hod}}\mu=0$, i.e. that $\mu$ is harmonic.
However, why does $\nabla\omega=0$ imply $\Delta_{\operatorname{Hod}}\omega=0$ for $\omega\in\Omega^p(M)$? (E.g. stated by Encyclopedia of Mathematics.)
Edit: Is there some geometric interpretation of this statement?
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 $\Delta\omega=(d\delta+\delta d)\omega=0$.