I will guess that the optimum occurs for $k$ isolated vertices and a complete graph on the other $n-k$ where $k=\lfloor\frac{n+1}5\rfloor.$ The same count occurs for $k$ vertices of degree $n-1$ and no other edges so the other $n-k$ have degree $k.$
Past that I have these observations:
- A graph $G$ and the complement $\bar G$ give the same value to the sum.
- If the maximum degree in an optimal $G$ is $\Delta$ then any degree $\Delta$ vertex is connected to any other. This is because connecting two such increases some of the $|\deg(x_i)-\deg(x_j)|$ but decreases none.
- Similarly two vertices with the minimum degree are non-adjacent.
- For the type of graph I defined above, the count is $k(n-k)(n-k-1)^3.$ The maximum over the reals occurs at $$k=\frac{3\,n-\sqrt {4\,{n}^{2}-n+1}-1}5\approx \frac{n}{5}-\frac3{20}.$$
As commented, the exponent of $3$ is relevant. Take the conjectured optimal case of a $K_{4t}$ and $t$ isolated vertices. Deleting one edge reduces $2t$ degree difference from $4t-1$ to $4t-2$ and increases $2(4t-2)$ other differences from $0$ to $1.$ If one is summing the square or cubes of the differences that is worse. But with exponent $1$ that is an improvement.
NOTE Based on limited calculations, The same things seem maximal if we replace the exponent of 3 by 2