2D case
Let a convex quadrilateral $Q= \operatorname{conv}\{A_1,A_1',A_2,A_2'\}$ have vertex pairs $$(A_1,A_1'),\quad (A_2,A_2'),$$ and define the “diagonal vectors” $$d_i = \overrightarrow{A_iA_i'} = A_i' - A_i \in \mathbb{R}^2.$$ Form the diagonal triangle $$T = \operatorname{conv}\{O,d_1,d_2\}, \qquad \operatorname{Area}(T) = \tfrac{1}{2}|\det(d_1,d_2)|.$$ The “shoelace formula” gives $$ \operatorname{Area}(Q) =\tfrac12\bigl|\det(A_1'-A_1,\,A_2'-A_2)\bigr| =\tfrac12|\det(d_1,d_2)| =\operatorname{Area}(T) $$ No symmetry assumptions are needed.
| $Q$ | $T$ |
|---|---|
 |
 |
3D case
Now take three vertex pairs $$ (A_1,A_1'),\ (A_2,A_2'),\ (A_3,A_3'), $$ and define $d_i=A_i'-A_i$.
Let $\mathcal{O} = \operatorname{conv}\{A_1,A_1',A_2,A_2',A_3,A_3'\}$.
If these six points form a polyhedron with 8 triangular faces, one vertex from each pair on every face, then the volume satisfies $$ \operatorname{Vol}(\mathcal O) =\frac{1}{6}\bigl|\det(d_1,d_2,d_3)\bigr| =\operatorname{Vol}(\operatorname{conv}\{O,d_1,d_2,d_3\}). $$
Sketch of proof:
Fix one “axis” edge $A_1A_1'$. The four faces incident to that edge are
$$
\triangle(A_1, A_2, A_3),\
\triangle(A_1, A_2', A_3),\
\triangle(A_1, A_2, A_3'),\
\triangle(A_1, A_2', A_3').
$$
They bound four tetrahedra
$T_{\varepsilon,\delta}=\operatorname{conv}\{A_1,A_1',A_2^{(\varepsilon)},A_3^{(\delta)}\}$
with $\varepsilon,\delta\in\{0,1\}$.
These tetrahedra fill $\mathcal O$ disjointly.
Hence $$ \operatorname{Vol}(\mathcal O) =\frac{1}{6}\sum_{\varepsilon,\delta}\bigl|\det(d_1,\,A_2^{(\varepsilon)}-A_1,\,A_3^{(\delta)}-A_1)\bigr|. $$ Dropping absolute values (Swapping $A_2$ and $A_2'$ in a term gives the reverse orientation) Using multilinearity in the last two slots, $$ \sum_{\varepsilon,\delta}\det(d_1,\,A_2^{(\varepsilon)}-A_1,\,A_3^{(\delta)}-A_1) =\det(d_1,\,A_2'-A_2,\,A_3'-A_3) =\det(d_1,d_2,d_3). $$ Thus $$ \operatorname{Vol}(\mathcal O)=\frac{1}{6}|\det(d_1,d_2,d_3)|. $$
Question
For higher dimensions:
Let $A_i,A_i'\in\mathbb{R}^n$ for $i=1,\dots,n$, and set $d_i=A_i'-A_i$. Define the $n$-dimensional “cross-polytope” $$ \mathcal{C}=\operatorname{conv}\{A_1,A_1',\dots,A_n,A_n'\}. $$ If each face of $\mathcal C$ uses exactly one vertex from each pair, is it true that $$ \operatorname{Vol}_n(\mathcal C) =\frac{1}{n!}\,|\det(d_1,\dots,d_n)|, $$ the same as the simplex spanned by the diagonal vectors?
