Since the Hodge star operation is a linear operation $*:\Lambda^kV \to \Lambda^{n-k} V$, then I thought that one could represent this linear map by tensor as one can can do for a linear map $A:V\to W$ by ${A^i}_j\ w_i\otimes v^j \in W\otimes V^*$, where $\{w_i\}$ resp. $\{v^j\}$ are basis of $W$ resp. dual basis of $V^*$.
So for an orthonormal basis $\{e_i\}$ of $V$ and and their algebraic dual $\{e^j\}$ in $V^*$, if we denote the basis $\{e_{i_1}\wedge\dots\wedge e_{i_k}\}$ of $\Lambda^k V$ as $\{e_I\}$, and similarly for $\Lambda^{n-k} V^* \cong (\Lambda^{n-k} V)^*$ as $\{e^J\}$, where $I$ resp. $J$ are increaing $k$ resp. $n-k$ indices from $\{1,\dots,n\}$, we should get $$ * = \sum_{I,J}\text{sgn}(I,J) \ e_I\otimes e^J \quad \in \quad(\Lambda^{n-k}V) \otimes (\Lambda^k V^*) $$ where the permutation $(I,J)$ is the one that sends the first $k$ indices from $\{1,\dots,n\}$ to $I$ and the rest to $J$. It is clear that the sum is only over $I$ and $J$ with $I\cap J = \emptyset$, or equivalently $I \cup J = \{1,\dots,n\}$.
In particular, for $k = 1$ we get the cross product of $n-1$ vectors in $\Lambda^1 V \cong V$, which lies in the space $\Lambda^{n-1}V$ $$ * = \sum_{j=1}^n(-1)^{j+1} e_j\otimes (e^1\wedge\dots\wedge\widehat{e^j}\wedge\dots\wedge e^n) $$ where a hat on an element means its absence. I hope that there is no mistake in this representation.