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
Let $p(v_1, \dots v_n)$ and $q(v_1, \dots v_n)$ be two parenthesizations of the cross product of $n$ vectors. First note that if $p$ and $q$ have any "bottom-level" cross products $v_i \times v_{i+1}$ in common (e.g. your examples have $C \times D$ in common) then since the cross product is surjective, $p = q$ iff $p = q$ with that bottom-level cross product replaced with a single vector (e.g. in your example we can replace $C \times D$ with just $C$). So we can assume WLOG that $p$ and $q$ have no bottom-level cross products in common.
Now let $v_i \times v_{i+1}$ be a bottom-level cross product in $p$, which by our WLOG assumption is not a bottom-level cross product in $q$. This means if we set $v_i = v_{i+1}$ then $p = 0$. Now it suffices to show that we can set the rest of the vectors $v_j$ such that $q \neq 0$.
Here's a sketch of how to do this (this should really include a tree diagram, my apologies). By the surjectivity of the cross product, we can WLOG replace any bottom-level cross product $v_j \times v_{j+1}$ in $q$ which does not involve either $v_i$ or $v_{i+1}$ with a single vector. After we perform this replacement as many times as possible, $q$ must only have two bottom-level cross products, namely $v_{i-1} \times v_i$ and $v_{i+1} \times v_{i+2}$. If we denote the standard basis of $\mathbb{R}^3$ by $e_1, e_2, e_3$ then we can set $v_i = v_{i+1} = e_1, v_{i-1} = -e_2, v_{i+2} = -e_3$ so that
$$v_{i-1} \times v_i = e_3, v_{i+1} \times v_{i+2} = e_2.$$
Now the structure of $q$ is that it takes the cross product of each of these with some other vectors, namely $v_{i-2}$ etc. on the left and $v_{i+3}$ etc. on the right, before eventually taking the cross product of the two resulting vectors, then taking some more cross products with some other vectors on the left and right. This last set of cross products is irrelevant; it suffices to show that the cross product of the subexpression involving $v_i$ and the subexpression involving $v_{i+1}$ can be taken to be nonzero. By induction we can always choose $v_j, j < i-1$ from the set $B = \{ \pm e_1, \pm e_2, \pm e_3 \}$ such that the subexpression involving $v_i$ is in $B$ by just repeatedly choosing such that the current cross product is nonzero, e.g. with the above choices so that $v_{i-1} \times v_i = e_3$ we could choose $v_{i-2} = e_2$ so that $v_{i-2} \times (v_{i-1} \times v_i) = e_1$; moreover at any step we always have two choices up to sign. The same is also true of the $v_j, j > i+1$ and of the subexpression involving $v_{i+1}$ which means that for both subexpressions, at the point where we make the final choice we have two choices on each side and so we can always guarantee that they are different up to sign, hence orthogonal, so that their cross product is a nonzero element of $B$.