The number of positive integer solutions to the equation $x_1+2x_2+...+nx_n=n^2.$

Note that $A_n$ is also equal to the number of tuples $(y_1,y_2,\ldots,y_n)\in\mathbb{Z}^n$ such that $0<y_1<y_2< \ldots y_n$ and $y_1+\ldots +y_n=n^2$.

[Consider the bijection $y_1=x_n,y_2=x_n+x_{n-1}$,$\ldots$,$y_n=x_n+\ldots +x_1$.]

The number of such tuples is less than $\dfrac{1}{n!}$ times the number of positive integer solutions to $z_1+\ldots +z_n=n^2$, which is ${n^2-1 \choose n-1}$, which is at most $\dfrac{n^{2(n-1)}}{(n-1)!}=\dfrac{n^{2n-1}}{n!}$—this gives the upper bound.

No time for the lower bound now.

From the OP's question The number of positive integer solutions to the equation $x_1+x_2+...+x_n=n^2.$, we have $$A_n>\frac{1}{2^{n-1}n!}\,\binom{n^2-1}{n-1}\geq\frac{\big(n(n-1)\big)^{n-1}}{2^{n-1}n!(n-1)!}=\frac{n^n(n-1)^{n-1}}{2^{n-1}(n!)^2}\,.$$