For $t=(t_1,\dots,t_n)$ in the cube $C_n$, let $A(t)=(a_{ij}(t))_{i,j=1}^{n+1}$ be the matrix whose entries $a_{ij}(t)$ are defined as follows:

$a_{ij}(t):=\partial_i f_j(t)$ for $i=1,\dots,n$ and $j=1,\dots,n+1$, where $\partial_k f_j(t)$ denotes the partial derivative of the function $f_j$ with respect to its $k$th argument $t_k$;

$a_{n+1,j}(t):=e_j$ for $j=1,\dots,n+1$, where $(e_1,\dots,e_{n+1})$ is the standard basis of $\mathbb R^{n+1}$.

In other words, for $i=1,\dots,n$, the $i$th row of the matrix $A(t)$ is the vector $\partial_i f(t):=(\partial_i f_1(t),\dots,\partial_i f_{n+1}(t))$, and the last row of $A(t)$ is $(e_1,\dots,e_{n+1})$.

So,
\begin{equation*}
\det A(t)=\bigwedge(\partial_1 f(t),\dots,\partial_n f(t)),
\end{equation*}
the generalized $n$-ary cross product of the vectors $\partial_1 f(t),\dots,\partial_n f(t)$ in $\mathbb R^{n+1}$; see e.g. Section Multilinear algebra.

It follows that the volume in question is

$$\int_{C_n}\|\det A(t)\|\,dt=\int_{C_n}\Big\|\bigwedge(\partial_1 f(t),\dots,\partial_n f(t))\Big\|\,dt,$$
where $\|\cdot\|$ is the Euclidean norm in $\mathbb R^{n+1}$.

The latter formula is an extension of the formula for the area of a 2D surface in $\mathbb R^3$ in terms of the cross product (in which case $n=2$), and its derivation in general is quite similar to that for $n=2$; cf. e.g. formulas (16.6.8)--(16.6.12). The key fact here is as follows. For any vectors $v_1,\dots,v_n$ in $\mathbb R^{n+1}$, let $B$ the matrix in $\mathbb R^{(n+1)\times(n+1)}$ with rows $v_1,\dots,v_{n+1}$, where $v_{n+1}:=\bigwedge(v_1,\dots,v_n)$. Note that the vector $v_{n+1}$ is orthogonal to the vectors $v_1,\dots,v_n$.
So, the $(n+1)$-volume $V_{n+1}(v_1,\dots,v_{n+1})$ of the parallelotope defined by the vectors $v_1,\dots,v_{n+1}$ is
\begin{equation*}
|\det B|=V_n(v_1,\dots,v_n)\|v_{n+1}\|; \tag{1}
\end{equation*}
here, $V_n(v_1,\dots,v_n)$ is of course the $n$-volume of the parallelotope defined by the vectors $v_1,\dots,v_n$.
On the other hand, expanding $\det B$ along its last row, we see that
\begin{equation*}
\det B=v_{n+1}\cdot\bigwedge(v_1,\dots,v_n)=\|v_{n+1}\|^2, \tag{2}
\end{equation*}
where $\cdot$ denotes the dot product.
Comparing (1) and (2), we conclude that
\begin{equation*}
V_n(v_1,\dots,v_n)=\|v_{n+1}\|=\Big\|\bigwedge(v_1,\dots,v_n)\Big\|.
\end{equation*}