We want to show:
$$*(e_{i_1} \wedge\ldots \wedge e_{i_k})=\epsilon e_{j_1} \wedge\ldots \wedge e_{j_{n-k}}$$
where $\epsilon$ is the sign of the permutation $i_1 \cdots i_k \bar i_1 \cdots \bar i_{n-k}.$
The book doesn't provide any short of proof but according to Wikipedia we look the Gram determinant of $$\det(e_{i_1} \wedge\ldots \wedge e_{i_k}\wedge e_{j_1} \wedge\ldots \wedge e_{j_{n-k}}).$$
I don't care so much about how we get the sign, but how we deduct the fact that $*(e_{i_1} \wedge\ldots \wedge e_{i_k})=\epsilon e_{j_1} \wedge\ldots \wedge e_{j_{n-k}}$ by looking the Gram determinant ?
In the comments you defined $*\beta$ indirectly by the property $$\forall \alpha\in\bigwedge^k \mathbb F^n: \alpha\wedge *\beta=\langle \alpha,\beta\rangle \textrm{vol}$$ where I expect that $\textrm{vol}$ is defined by $\textrm{vol}=e_1\wedge\ldots\wedge e_n$ (the unpermuted basis - note that the indices run from 1 to n, not from $i_1$ to $i_n$). Note that there is no choice of $\alpha$ to state this definition - $\alpha$ is not a free (independent) variable, it is bound by the $\forall$ quantifier. Therefore, when applying it, we are free to plug in whatever $\alpha$s we want.
Take $\alpha = \beta = e_{i_1} \wedge\ldots \wedge e_{i_k}$. Then $\langle \alpha,\beta\rangle =1$ and $$e_{i_1}\wedge\ldots\wedge e_{i_k}\wedge *\beta = \alpha\wedge*\beta=\textrm{vol}.$$
Note that the left-hand side uses the permuted basis, so to extract $*\beta$ from this expression we have to write $\textrm{vol}$ (the RHS) in terms of the permuted basis, too. By permuting terms in $e_1\wedge\ldots\wedge e_n$ (with the permutation $i:\{1...n\}\to\{1...n\}$) we get $\epsilon \;e_{i_1} \wedge\ldots \wedge e_{i_n}$, where the constant factor $\epsilon$ is the sign of the permutation $i$. It pops up because of the antisymmetry of the wedge product. Therefore,
$$e_{i_1}\wedge\ldots\wedge e_{i_k}\wedge *\beta = \epsilon \;e_{i_1} \wedge\ldots \wedge e_{i_n}.$$
Now we wish to equate the rightmost $n-k$ terms to get the desired result $*\beta=\epsilon\;e_{i_{k+1}}\wedge\ldots\wedge e_{i_n}$.
Unfortunately this doesn't work unless we prove $*\beta$ has no summands containing any of $e_{i_1},\ldots,e_{i_k}$. That is, right now we only know $$*\beta\in\epsilon\;e_{i_{k+1}}\wedge\ldots\wedge e_{i_n} + \textrm{span} \{e_{\sigma_1}\wedge \ldots\wedge e_{\sigma_{n-k}} : \sigma_j=i_j\textrm{ for some } j \}$$
(It may be easier to imagine it in 2D: $e_1\wedge \xi=e_1\wedge e_2$ does not necessarily imply $\xi=e_2$, it only implies $\xi\in e_2+\textrm{span}\{e_1\}$).
We can easily rule out the excess span though, by applying $\alpha\wedge*\beta=\langle\alpha,\beta\rangle\textrm{vol}$ with $\alpha$ ranging over its basis:
$$\langle e_{\sigma_1}\wedge \ldots\wedge e_{\sigma_{n-k}}, \beta\rangle = 0 \textrm{ if } \sigma_j=i_j\textrm{ for some } j$$ holds for any permutation $\sigma$ because a scalar product between multivectors consisting of different basis vectors is zero (just like in 3D $\langle e_1\wedge e_2, e_2\wedge e_3\rangle=0$). Therefore the excess span in our conclusion about $*\beta$ plays no role and $*\beta$ is just $\epsilon\;e_{i_{k+1}}\wedge\ldots\wedge e_{i_n}$.
You may find it interesting that this is also a proof of existence of $*\beta$. The definition of $*\beta$ you took only states a property of an imaginary object $*\beta$. The statement we prove here gives a formula for it over the basis of $\bigwedge^k \mathbb F^n$ (we did this by taking $\beta=e_{i_1} \wedge\ldots \wedge e_{i_k}$ over any permutation $i$), and by linear extension $*$ becomes a linear operator $*:\bigwedge^k \mathbb F^n\to \bigwedge^{n-k} \mathbb F^n$.
Since the defining property of $*\beta$ depends linearly on $\beta$ (in the second place of the scalar product), we confirm that the linearly extended $*$ satisfies the defining property over all of $\bigwedge^k \mathbb F^n$ (and not just the basis vectors), thus proving existence of $*\beta$ for all $\beta$.