If we parametrize the plane $P:x+y+z=0$ (bijectively) using
$$
\matrix{
u&=&\frac{x+y}{\sqrt2}\\
v&=&\frac{x-y}{\sqrt2}
}
\qquad
\text{as}
\qquad
\matrix{
x&=&\frac{u+v}{\sqrt2}\\
y&=&\frac{u-v}{\sqrt2}\\
z&=&-\sqrt2\,u
}
$$
then the orthogonal projection of the hexagon $H$
at the intersection of the cube $C$
$$\max\Bigl(|x|,|y|,|z|\Bigr)=\frac12$$
onto $P$ can be found by simply describing $(u,v)$ along $H$.
(In fact $H$ is already in the plane $P$.)
In particular, if a symmetry about $(0,0)$ can be observed,
then we will have demonstrated that the centers of $C$ and $H$ coincide.
Using the given parametrization actually already guarantees that $(u,v)$
that $(x,y,z)$ lies in the plane $P$,
so we need only transform the equation for $C$:
$$\max\left(
\left|\frac{u+v}{\sqrt2}\right|,
\left|\frac{u-v}{\sqrt2}\right|,
\left|\sqrt2\,u\right|
\right)=\frac12$$
$$\max\left(
\left|u+v\right|,
\left|u-v\right|,
\left|2\,u\right|
\right)=\frac1{\sqrt2}$$
This can be visualized in the $uv$-plane
as the hexagonal curve with the origin at its interior
connecting the finite line segments
of the six lines with equations
$$
\matrix{
u+v&=&\pm s \\\\
u-v&=&\pm s \\\\
u&=&\pm\frac{s}{2}
}
$$
for $s=\frac1{\sqrt2}$.
It's pretty easy to see that its center is $(0,0)$.
The three pairs of equations above are plotted
below in red, green and blue, respectively,
and the projections of the portions of the
$xzy$-coordinate axes interior to $C$
(between the centers of opposite faces)
is shown dotted in gray.
