Given $n+1$ points, bound the product of the distances from one of them

Yes, it is true. Consider the polynomial $p(x)=\prod_{i=1}^{n+1}(x-x_i)$ and the monic Chebyshev polynomial $T_n(x)=x^n+\dots$ of degree $n$, so $|T_n|\le 2^{-(n-1)}$ on $[-1,1]$. Now apply the residue theorem to the integral of the rational function $Q(z)=\frac{T_n(z)}{p(z)}$ in a huge disk centered at $0$. On the one hand $$ \oint_{|z|=R}Q(z)\,dz\approx \oint_{|z|=R}\frac{dz}{z}=2\pi i $$ as $R\to\infty$ because $Q(z)=\frac 1z+O(|z|^{-2})$ as $z\to\infty$.

On the other hand, it is $$2\pi i\sum_j{\rm Res}_{x_j}Q=2\pi i\sum_j\frac{T_n(x_j)}{p'(x_j)}\,.$$ Since $|T_n(x_j)|\le 2^{-(n-1)}$, we conclude that $$\sum_j\frac{1}{|p'(x_j)|}\ge 2^{n-1}\,,$$ so there must exist $j$ with $|p'(x_j)|\le \frac{n+1}{2^{n-1}}$. But $|p'(x_j)|$ are exactly the products you are interested in.