There's a simple tool for this: First, by translating, we may assume $\newcommand{\bfe}{{\bf e}} \newcommand{\bfx}{{\bf x}} \bfx_0 = {\bf 0}$. The $k$-dimensional volume $\left|\left| \bfx_1 \wedge \cdots \wedge \bfx_k \right|\right|$ of the parallelotope determined by the vectors $ \bfx_1, \ldots, \bfx_k$ is given by the square root of the Gram determinant, that is, the determinant of the Gram matrix
$$ \pmatrix{\bfx_1^T \\ \vdots \\ \bfx_k^T} \pmatrix{\bfx_1 & \cdots & \bfx_k}
=
\pmatrix{
\langle \bfx_1, \bfx_1 \rangle & \cdots & \langle \bfx_1, \bfx_k \rangle \\
\vdots & & \vdots \\
\langle \bfx_k, \bfx_1 \rangle & \cdots & \langle \bfx_k, \bfx_k \rangle
} .
$$
The volume of the $k$-simplex $S$ determined by $\bfx_1 , \ldots , \bfx_k$ is a fraction of this:
$$\color{#bf0000}{\boxed{\operatorname{vol}(S) = \frac{1}{k!} \left|\left| \bfx_1 \wedge \cdots \wedge \bfx_k \right|\right| }}.$$ (One can verify directly that $\tfrac{1}{k!}$ is the right coefficient by computing the volume of the standard $k$-simplex with vertices $\bfe_1, \ldots, \bfe_k$, where $(\bfe_a)$ is the standard basis of $\Bbb R^n$.)
In the special case $k = n - 1$, $\left|\left| \bfx_1 \wedge \cdots \wedge \bfx_{n - 1} \right|\right|$ is the length of the vector $\ast (\bfx_1 \wedge \cdots \wedge \bfx_{n - 1})$ given by formally evaluating the determinant
$$
\det \pmatrix{
x_{11} & \cdots & x_{1n} \\
\vdots & & \vdots \\
x_{n-1, 1} & \cdots & x_{n - 1, n} \\
\bfe_1 & \cdots & \bfe_n
} .
$$
For the case $n = 3$, this formal determinant is just the usual cross product of vectors in $\Bbb R^3$, and indeed, the map $(\bfx_1, \ldots, \bfx_{n - 1}) \mapsto \ast (\bfx_1 \wedge \cdots \wedge \bfx_{n - 1})$ is sometimes called an $(n - 1)$-fold cross product.
