The minimum must occur at vectors $x,y$ where
$x_i$ and $y_i$ take only two values each.
This should make it easy to check **Neil Strickland**'s
experimental result (where $x$ and $y$ are indeed of this form).

It is convenient to extend $A$ homogeneously to all nonzero $x,y$
in the zero-sum hyperplane:
$$
A(x,y) = \frac1{\| x \| \, \|y\|} \sum_{1\leq i<j\leq n} |x_i-x_j|\ |y_i-y_j|
$$
where $\|x\| = (x_1^2 + \cdots + x_n^2)^{1/2}$ and likewise $\|y\|$.
So we seek the largest $a$ such that
$$
(1)\qquad\qquad\qquad
\sum_{1\leq i<j\leq n} |x_i-x_j|\ |y_i-y_j| \geq a \, \|x\| \, \|y\|
\qquad\qquad\qquad\phantom{(1)}
$$
for all $x,y$ such that $\sum_{i=1}^n x_i = \sum_{i=1}^n y_i = 0$.

Fix one of the $n!^2$ possible orders of the coordinates of $x$ and $y$.
Such a choice limits each of $x$ and $y$ to a cone whose extreme rays are
generated by $(1,\ldots,1,1-n)$, $(2,\ldots,2,2-n,2-n)$, etc., all with
$x_i$ or $y_j$ take only two values each.
Given that choice of order, $\sum_{1\leq i<j\leq n} |x_i-x_j| |y_i-y_j|$
is a bilinear form in $x,y$ because the signs of $x_i-x_j$ and $y_i-y_j$
are constant.  The claim then follows by a convexity argument
(so ultimately by the triangle inequality).  Indeed if we fix $x$,
and (1) holds for $(x,y)$ and $(x,y')$ for some $y,y'$ in the same cone,
then it also holds for $(x, ty+(1-t)y')$ for all $t \in [0,1]$.
So it is enough to check (1) for $y$ on an extreme ray of the cone.
Likewise we can fix $y$ and reduce to the case where $x$ is on an extreme ray.
**QED**