Consider Euclidean 2-space, with line element \begin{equation} ds^2 = dx^2 + dy^2 \end{equation} and (ordered) orthonormal basis $\{dx,dy\}$. The orientation is \begin{equation} \omega = dx\wedge dy \end{equation} and it is again straightforward to compute the Hodge dual on a basis. In analogy with the previous example, we have \begin{equation} dx \wedge {*}dx = g(dx,dx)\, dx \wedge dy = dx \wedge dy \end{equation} from which it follows that \begin{equation} {*}dx = dy \end{equation} Similarly, from \begin{equation} dy \wedge {*}dy = g(dy,dy)\, dx \wedge dy = dx\wedge dy \end{equation} we see that \begin{equation} {*}dy = -dx \end{equation} As before, the remaining cases were already worked out in general; for Euclidean 2-space we obtain \begin{align} {*}1 &= dx\wedge dy \\ {*}(dx\wedge dy) &= 1 \end{align}

It is important to note that, even in this positive-definite setting, there are some factors of $-1$ in these relations, which arise from the necessary permutations. You should verify for yourself that no such factors arise in Euclidean 3-space.