On every Riemannian manifold $M$, we can consider the Hodge $*$-operator, which is characterised by the following formula: $$a \wedge *b = (a,b)\nu.$$ Here $a$ and $b$ are smooth forms on $M$, $(\ ,\ )$ is a metric on $\wedge T^*\!M$ and $\nu$ is the volume form with respect to the Riemannian metric.
My question: Is a formula of $*(a \wedge b)$ known?
I suspect that we can have a formula like "$*(a \wedge b)=(*a)\wedge(*b)$" or "$*(a \wedge b) = *a \wedge b \pm a \wedge *b$". Of course these formulae never hold. (Look at the degree.)
A toy example. Given an orthonormal basis $e_1,\dots,e_n$ for the vector space $V$, then $\nu=e_1\wedge\dots\wedge e_n$ and
$$*(e_1\wedge e_2)=e_3\wedge\dots\wedge e_n\in \bigwedge^{n-2}V, $$
with $*e_1=e_2\wedge\dots\wedge e_n$ and $*e_2=-e_1\wedge e_3\wedge\dots\wedge e_n$ both in $\bigwedge^{n-1}V$. Introducing the insertion operator $i_\bullet: \bigwedge^{k}V\rightarrow \bigwedge^{k-1}V$ we arrive at
$$*(e_1\wedge e_2)=\frac{1}{2}\left( \underbrace{i_{e_2}(\underbrace{*e_1}_{\in \bigwedge^{n-1}V}}_{\in \bigwedge^{n-2}V})-i_{e_1}(*e_2)\right). $$
I do not if this is what you were searching for, but I hope it helps (a bit).