If $x_1,\dotsc,x_{n-1} \in \mathbb{R}^n$, one defines $x_1 \times \cdots \times x_{n-1} \in \mathbb{R}^n$ to be the unique vector such that
$$
\forall y \in \mathbb{R}^n, \quad \langle x_1 \times \cdots \times x_{n-1},y \rangle = \operatorname{det}(x_1,\dotsc,x_{n-1},y),
$$
where the determinant is being viewed as a function of the rows or columns of the usual matrix argument, i.e., as the unique antisymmetric $n$-form $\operatorname{det} : \mathbb{R}^n \times \cdots \times \mathbb{R}^n \to \mathbb{R}$ such that $\det(e_1,\dotsc,e_n) = 1$ for $\{e_k\}$ the standard ordered basis of $\mathbb{R}^n$.
Now, suppose that $x_1,\dotsc,x_{n-1} \in \mathbb{R}^n$ are linearly independent, and hence span a hyperplane $H$ ($n-1$-dimensional subspace) in $\mathbb{R}^n$. Then, in particular, $x_1 \times \cdots \times x_{n-1} \neq 0$ is orthogonal to each $x_k$, and hence defines a non-zero normal vector to $H$; write $$x_1 \times \cdots \times x_{n-1} = \|x_1 \times \cdots \times x_{n-1}\|\hat{n}$$ for $\hat{n}$ the corresponding unit normal. Let $y \notin H$. Then $x_1,\dotsc,x_{n-1},y$ are linearly independent and span an $n$-dimensional parallelopiped $P$ with $n$-dimensional volume
$$
|\operatorname{det}(x_1,\dotsc,x_{n-1},y)| = |\langle x_1 \times \cdots x_{n-1},y\rangle| = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|.
$$
Now, with respect to the decomposition $\mathbb{R}^n = H^\perp \oplus H$, let
$$
T = \begin{pmatrix} I_{H^\perp} & 0 \\ M & I_{H} \end{pmatrix}
$$
for $M : H^\perp \to H$ given by $$M(c \hat{n}) = -c \langle \hat{n},y \rangle^{-1} P_H y = -c\langle \hat{n},y\rangle^{-1}(y-\langle\hat{n},y\rangle\hat{n}),$$ where $P_H(v)$ denotes the orthogonal projection of $v$ onto $H$. Then $T(P)$ is a $n$-dimensional parallelepiped with with vertices $Tx_1 = x_1,\dotsc,Tx_{n-1}=x_{n-1}$, and
$$
Ty = \langle \hat{n},y \rangle \hat{n} = P_{H^\perp} y = y - P_H y,
$$
with the same volume as $P$. On the one hand, since $Ty = y - P_H y$ for $P_H y \in H = \{x_1 \times \cdots \times x_{n-1}\}^\perp$,
$$
\operatorname{Vol}_n(T(P)) = |\operatorname{det}(Tx_1,\dotsc,Tx_{n-1},Ty)|\\ = |\operatorname{det}(x_1,\dotsc,x_{n-1},y-P_H y)|\\ = |\operatorname{det}(x_1,\dotsc,x_{n-1},y)|\\ = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|.
$$
On the other hand, since $Ty \in H^\perp$, $T(P)$ is an honest cylinder with height $\|Ty\| = |\langle \hat{n},y\rangle|$ and base the $(n-1)$-dimensional parallelopiped $R$ spanned by $x_1,\dotsc,x_{n-1}$, so that
$$
\operatorname{Vol}_n(T(P)) = \operatorname{Vol}_{n-1}(R)|\langle \hat{n},y\rangle|.
$$
Thus,
$$
\operatorname{Vol}_{n-1}(R)|\langle \hat{n},y\rangle| = \operatorname{Vol}_n(T(P)) = \|x_1 \times \cdots \times x_{n-1}\||\langle \hat{n},y\rangle|,
$$
so that
$$
\operatorname{Vol}_{n-1}(R)| = \|x_1 \times \cdots \times x_{n-1}\|,
$$
as required.
EDIT: Theoretical Addendum
Let's see what $\phi x_1 \times \cdots \times \phi x_n$ is in terms of $x_1 \times \cdots \times x_{n-1}$ for $\phi$ a linear transformation on $\mathbb{R}^n$.
Define a linear map $T : (\mathbb{R}^n)^{\otimes(n-1)} \to (\mathbb{R}^n)^\ast$ by
$$
T : x_1 \otimes \cdots \otimes x_{n-1} \mapsto \operatorname{det}(x_1,\cdots,x_{n-1},\bullet),
$$
so that if $S : \mathbb{R}^n \to (\mathbb{R}^n)^\ast$ is the isomorphism $v \mapsto \langle v,\bullet \rangle$, then
$$
x_1 \times \cdots \times x_n = (S^{-1}T)(x_1 \otimes \cdots \otimes x_n).
$$
Now, since the determinant is antisymmetric, so too is $T$, and hence $T$ descends to a linear map $T : \bigwedge^{n-1} \mathbb{R}^n \to (\mathbb{R}^n)^\ast$,
$$
x_1 \wedge \cdots \wedge x_{n-1} \mapsto \operatorname{det}(x_1,\cdots,x_{n-1},\bullet);
$$
indeed, if $\operatorname{Vol} = e_1 \wedge \cdots \wedge e_n$ for $\{e_k\}$ the standard ordered basis for $\mathbb{R}^n$, then for any $y \in \mathbb{R}^n$,
$$
\langle x_1 \otimes \cdots \otimes x_{n-1},y \rangle \operatorname{Vol} = \operatorname{det}(x_1,\cdots,x_{n-1},y)\operatorname{Vol} = x_1 \wedge \cdots \wedge x_{n-1} \wedge y,
$$
which, in fact, shows that
$$
x_1 \times \cdots \times x_{n-1} = \ast (x_1 \wedge \cdots \wedge x_{n-1}),
$$
where $\ast : \wedge^{n-1} \mathbb{R}^n \to \mathbb{R}^n$ is the relevant Hodge $\ast$-operator. Thus, a cross product is really an $(n-1)$-form in the orientation-dependent disguise given by the Hodge $\ast$-operator; in particular, it will really transform as an $(n-1)$-form, as we'll see now.
Now, let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be linear. Observe that the adjugate matrix $\operatorname{Adj}(\phi)$ of $\phi$ can be invariantly defined as the unique linear transformation $\operatorname{Adj}(\phi) : \mathbb{R}^n \to \mathbb{R}^n$ such that for any $\omega \in \bigwedge^{n-1} \mathbb{R}^n$ and $y \in \mathbb{R}^n$,
$$
(\wedge^{n-1})\omega \wedge y = \omega \wedge \operatorname{Adj}(\phi) y,
$$
e.g., in our case,
$$
x_1 \wedge \cdots \wedge x_{n-1} \wedge \operatorname{Adj}(\phi) y = (\wedge^{n-1}\phi)(x_1 \wedge \cdots \wedge x_{n-1}) \wedge y = \phi x_1 \wedge \cdots \wedge \phi x_{n-1} \wedge y,
$$
and that, as a matrix, $\operatorname{Adj}(\phi) = \operatorname{Cof}(\phi)^T$, where $\operatorname{Cof}(\phi)$ denotes the cofactor matrix of $\phi$. Then for any $y$,
$$
\langle \phi x_1 \times \cdots \times \phi x_{n-1},y \rangle \operatorname{Vol} = \operatorname{det}(\phi x_1,\cdots,\phi x_{n-1},y)\operatorname{Vol}\\ = \phi x_1 \wedge \cdots \wedge \phi x_{n-1} \wedge y\\ = (\wedge^{n-1}\phi)(x_1 \wedge \cdots \wedge x_{n-1}) \wedge y\\ = (x_1 \wedge \cdots \wedge x_{n-1}) \wedge \operatorname{Adj}(\phi)y\\ = \langle x_1 \times \cdots \times x_{n-1},\operatorname{Adj}(\phi)y \rangle \operatorname{Vol}\\ = \langle \operatorname{Cof}(\phi)(x_1 \times \cdots \times x_{n-1}),y \rangle \operatorname{Vol},
$$
and hence, since $y$ was arbitrary,
$$
\phi x_1 \times \cdots \times \phi x_{n-1} = \operatorname{Cof}(\phi)(x_1 \times \cdots \times x_{n-1}) = (\ast \circ \wedge^{n-1}\phi \circ \ast^{-1})(x_1 \times \cdots \times x_{n-1}),
$$
in terms of the Hodge $\ast$-operation and the invariantly defined $\wedge^{n-1}\phi$.
Best Answer
Yes. It is just like in dimension $3$: if your vectors are $(t_1,t_2,t_3,t_4)$, $(u_1,u_2,u_3,u_4)$, and $(v_1,v_2,v_3,v_4)$, compute the formal determinant:$$\begin{vmatrix}t_1&t_2&t_3&t_4\\u_1&u_2&u_3&u_4\\v_1&v_2&v_3&v_4\\e_1&e_2&e_3&e_4\end{vmatrix}.$$ You then see $(e_1,e_2,e_3,e_4)$ as the canonical basis of $\mathbb{R}^4$. Then the previous determinant is $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ with\begin{align*}\alpha_1&=t_4u_3v_2-t_3u_4v_2-t_4u_2v_3+t_2u_4v_3+t_3u_2v_4-t_2u_3v_4\\\alpha_2&=-t_4u_3v_1+t_3u_4v_1+t_4u_1v_3-t_1u_4v_3-t_3u_1v_4+t_1u_3v_4\\\alpha_3&=t_4u_2v_1-t_2u_4v_1-t_4u_1v_2+t_1u_4v_2+t_2u_1v_4-t_1u_2v_4\\\alpha_4&=-t_3u_2v_1+t_2u_3v_1+t_3u_1v_2-t_1u_3v_2-t_2u_1v_3+t_1u_2v_3\end{align*}It's a vector orthogonal to the other three.
I followed a suggestion taken from the comments on this answer: to put the entries $e_1$, $e_2$, $e_3$, and $e_4$ at the bottom. It makes no difference in odd dimension, but it produces the natural sign in even dimension.
Following another suggestion, I would like to add this remark:$$\alpha_1=-\begin{vmatrix}t_2&t_3&t_4\\u_2&u_3&u_4\\v_2&v_3&v_4\end{vmatrix}\text{, }\alpha_2=\begin{vmatrix}t_1&t_3&t_4\\u_1&u_3&u_4\\v_1&v_3&v_4\end{vmatrix}\text{, }\alpha_3=-\begin{vmatrix}t_1&t_2&t_4\\u_1&u_2&u_4\\v_1&v_2&v_4\end{vmatrix}\text{ and }\alpha_4=\begin{vmatrix}t_1&t_2&t_3\\u_1&u_2&u_3\\v_1&v_2&v_3\\\end{vmatrix}.$$