**EDIT 2018-04-16**: Formulae are corrected.

I'm not sure about connection with $\beta_{mv}$, but we can obtain a recurrence formula for $A_{j,m}$ as follows.

First let us fix the unused values $A_{j,m}=0$ for $j<0$ or $j>m$, so we won't need to care about the summation range for $j$.

Expanding $(n-k)^j$ and using Faulhaber's formula, we get
$$\sum_{k=0}^{n-1} (n-k)^j k^j = \sum_{k=0}^{n-1} \sum_i \binom{j}i n^{j-i} (-1)^i k^{i+j}$$
$$=\sum_{i} \binom{j}{i} n^{j-i} \frac{(-1)^i}{i+j+1} \left[ \sum_t \binom{i+j+1}t B_t n^{i+j+1-t} - B_{i+j+1}\right]$$
$$=\sum_{i,t} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t B_t n^{2j+1-t} - \sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} B_{i+j+1} n^{j-i}$$
where $B_t$ are Bernoulli numbers.

Now, we notice that
$$\sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t
=\begin{cases}
\frac{1}{(2j+1)\binom{2j}j}, & \text{if } t=0;\\
\frac{(-1)^j}{t}\binom{j}{2j-t+1}, & \text{if } t>0.
\end{cases}
$$
In particular, the last sum is zero for $0<t\leq j$.

Hence, introducing $\ell=2j+1-t$ and $\ell=j-i$, respectively, we get
$$\sum_{k=0}^{n-1} (n-k)^j k^j
= \frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + \sum_{\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell} - \sum_{\ell} \binom{j}{\ell} \frac{(-1)^{j-\ell}}{2j+1-\ell} B_{2j+1-\ell}n^{\ell}$$
$$=\frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + 2\sum_{\text{odd }\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell}.$$

Using the definition of $A_{j,m}$, we obtain the following identity for polynomials in $n$:
$$(\star)\qquad\sum_{j} A_{j,m} \frac{1}{(2j+1)\binom{2j}j}n^{2j+1}
+ 2 \sum_{j,\text{ odd }\ell} A_{j,m} \binom{j}{\ell} \frac{(-1)^j}{2j+1-\ell} B_{2j+1-\ell}n^{\ell} \equiv n^{2m+1}.$$

Taking the coefficient of $n^{2m+1}$ in $(\star)$, we get $A_{m,m} = (2m+1) \binom{2m}{m},$
and taking the coefficient of $x^{2d+1}$ for an integer $d$ in the range $m/2 \leq d < m$, we get $A_{d,m} = 0$.

Taking the coefficient of $n^{2d+1}$ in $(\star)$ for $m/4 \leq d < m/2$, we get
$$A_{d,m} \frac{1}{(2d+1)\binom{2d}{d}} + 2 (2m+1) \binom{2m}{m} \binom{m}{2d+1} \frac{(-1)^m}{2m-2d} B_{2m-2d} = 0,$$
i.e.
$$A_{d,m} = (-1)^{m-1} \frac{(2m+1)!}{d!d!m!(m-2d-1)!}\frac{1}{m-d} B_{2m-2d}.$$

Continue similarly, we can express $A_{d,m}$ for each integer $d$ in the range $m/2^{s+1}\leq d< m/2^s$ (iterating consecutively $s=1,2,\dots$) via previously determined values of $A_{j,m}$ as follows:
$$A_{d,m} = (2d+1)\binom{2d}{d} \sum_{j\geq 2d+1} A_{j,m} \binom{j}{2d+1} \frac{(-1)^{j-1}}{j-d} B_{2j-2d}.$$
The same formula holds also for $d=0$.