The numbers $a_{n,m}$ are in fact the Fourier coefficients of the polynomial
$$P_n(x)=\prod_{j=1}^{n-1} \Big( \frac{nx}{2} + \frac{n}{2}-j\Big) $$
with respect to the Chebyshev measure $d\sigma:=(1-x^2)^{-1/2}dx$ on $[-1,1]$, and its orthogonal bases of the Chebyshev polynomials of the first kind. Precisely,
for $0\le m\le n$
$$a_{n,m}=\frac{1}{\pi}  \int_{-1}^1 P_n(x)T_m(x)d\sigma  \ .$$

Changing variable, we have a trigonometric version:
$$a_{n,m}=\frac{1}{\pi}\int_0^\pi P_n(\cos \theta)\cos (m\theta)d\theta \ .$$


Note that the polynomials  $P_n$ and $T_m$ are odd resp. even, according to the parity of $n-1$, respectively $m$, so the integrand $P_n(x)T_m(x)$ has the same parity of $n+m-1$. On the other hand, the Chebyshev measure is symmetric, which explains the vanishing property $a_{n,m}=0$ whenever $n+m$ is even. Moreover, for odd $n+m$ the integrand is positive and concentrated about $\pm 1$; this should hopefully yield to the desired estimate $a_{n,m}>0$. I'll try some computation and in case add details later.

To compute the integral we may also use the Chebyshev-Gauss Quadrature formula on $N$ nodes, which is exact on polynomials of degree less than $2N$. Therefore, for  $2N\ge n+m$ we have
$$a_{n,m}=\frac{1}{N}\sum_{k=1}^{N} P_n\Big(\cos\big( \frac{2k-1}{2N}\pi \big) \Big)\ \cos\big( m\frac{2k-1}{2N}\pi\big) \ .$$