Your suggestion for $V^{*}$ is indeed a natural one and often used. A metric on $V$ gives you an isomorphism $T\colon V \rightarrow V^{*}$ (using Riesz Representation Theorem) and then you put an inner product on $V^{*}$ which turns $\phi$ into an isometry (so you define $\left< \varphi, \psi\right>_{V^{*}} := \left< T^{-1}(\varphi), T^{-1}(\psi) \right>_V$). Note that this means that if $(e_1,\dots,e_n)$ is an orthonormal basis then $(e^1 = T(e_1), \dots, e^n = T(e_n))$ is the dual basis of $(e_1,\dots,e_n)$ and so we see that the inner product on $V^{*}$ makes $e^1,\dots,e^n$ into an orthonormal basis of $V^{*}$.
Regarding the exterior powers $\Lambda^k(V)$, a natural choice is to define an inner product on decomposable $k$-wedges by the formula
$$ \left< v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k \right>_{\Lambda^k(V)} := \det \left( \left< v_i, w_j \right> \right)_{i,j=1}^k $$
and then to extend using bilinearity to all $\Lambda^k(V)$. The reason this is natural is that if $(e_1,\dots,e_n)$ is an orthonormal basis for $V$, then $(e_{i_1} \wedge \dots \wedge e_{i_k})_{I = (i_1 < \dots < i_k)}$ is an orthonormal basis for $\Lambda^k(V)$.