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

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). [EDIT: see the answer by Adam P. Goucher for this check.]

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. For the order $x_1 \geq x_2 \geq \ldots \geq x_n$, the cone's extreme rays are generated by $(1,\ldots,1,1-n)$, $(2,\ldots,2,2-n,2-n)$, etc., all with $x_i$ or $y_j$ taking only two values each. In general we have the image of these generators under some coordinate permutation, so still with $x_i$ or $y_j$ taking 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