In these contexts I often find it useful to work in an orthonormal basis. Let $e^1, \dots, e^n$ be an orthonormal basis for $T_x^\ast M$ (this is just a pointwise linear algebra computation, so we can fix a point $x \in M$). Note that $\sqrt{g} = 1$, so $\text{vol}_g = e^1 \wedge \dots \wedge e^n$. The key thing is that if we start with an orthonormal basis for $T_x^\ast M$, the set of forms of the form $e^{i_1} \wedge \dots \wedge e^{i_p}$, where $1 \leq i_1 < \dots < i_p \leq n$, form an orthonormal basis for $\Lambda^p T_x^\ast M$. From this it is immediate that $g(\text{vol}_g, \text{vol}_g) = 1$. Do you see how this implies that $\ast \text{vol}_g = 1$?
I now figured out something, but I am not sure whether my calculations are correct, so please tell me if my approach is acceptable.
Let $\alpha \in \Lambda^p$ such that $\alpha = \frac{1}{p!} \; \alpha_{k_1 \ldots k_p} \; dx^{k_1} \wedge \ldots dx^{k_p}$ and let $\beta = dx^{j_1} \wedge \ldots \wedge dx^{j_p}$. Then by assumption
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha \wedge\Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big) \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha_{k_1 \ldots k_p} \; \Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge \Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big)
\end{align}
By renaming the indices, one obtains
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \underbrace{\Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge\Big(dx^{k_{p+1}} \wedge \ldots \wedge dx^{k_n} \Big)}_{= \; \varepsilon^{k_1 \ldots k_n} \; dx^1 \wedge \ldots \wedge dx^n} \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \; \varepsilon^{k_1 \ldots k_n} \; \Big( dx^1 \wedge \ldots \wedge dx^n \Big)
\end{align}
Using $dV =\sqrt{\det g} \cdot dx^1 \wedge \ldots \wedge dx^n$, one gets
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \varepsilon^{k_1 \ldots k_n} \; \alpha_{k_1 \ldots k_p}\\
&= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; (n-p)! \; \delta^{k_1 \ldots k_p}_{j_1 \ldots j_p} \; \alpha_{k_1 \ldots k_p} \\
&= \enspace \frac{dV}{p!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; p! \; \alpha_{[j_1 \ldots j_p]} \\
&= \enspace dV \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \alpha_{j_1 \ldots j_p}
\end{align}
which coincides with the inner product. Here, the antiymmetry of $\alpha$ was used. Note that the factor of $\tfrac{1}{p!}$ within the inner product arises from a $p$-form $\gamma = \tfrac{1}{p!} \, \gamma_{m_1 \ldots m_p} \, dx^{m_1} \wedge \ldots \wedge dx^{m_p}$ being inserted into the inner product, i.e. $(\alpha, \gamma)$.
Please tell me what you think.
Best Answer
This appears to be a typo. (Indeed, if $\dim V$ is even and $\omega$ and $\eta$ are odd-degree forms, then $\ast \eta$ has odd degree, too, and $\omega \wedge \ast \eta = - \ast \eta \wedge \omega$.)
The identity in Theorem 1.3.2.1 should instead read $$\color{#bf0000}{\boxed{\eta \wedge \ast \omega = \langle \eta, \omega \rangle_{\Lambda^k V} dV = \langle \omega, \eta \rangle_{\Lambda^k V} dV = \omega \wedge \ast \eta}} .$$ The first and third equalities follow from the characterization of $\ast$ that you mention, and the middle equality is just the symmetry of the induced bilinear form $\langle \,\cdot\, , \,\cdot \, \rangle_{\Lambda^k V}$.