Finding a closed form for $I_n=\int_0^1 \prod_{l=1}^n\left[x^2-\frac{l^2}{n^2}\right]dx$

$$I_n=\frac{1}{n^{2n}}\int_0^1\prod_{l=1}^{n}[(xn)^2-l^2]dx\qquad x=\frac{y}{n}$$ $$I_n=\frac{1}{n^{2n+1}}\int_0^n\prod_{l=1}^{n}[y^2-l^2]dy$$ We can expand the product, without knowing the coefficients for now as following; $$\prod_{l=1}^{n}[y^2-l^2]=\sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$ $$I_n=\frac{1}{n^{2n+1}}\sum_{j=0}^n(-1)^{n-j}t(n,j)\int_0^ny^{2j}dy$$ $$I_n=\frac{1}{n^{2n+1}}\sum_{j=0}^n\frac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$ $$I_n=\frac{(-1)^n}{n^{2n}}\sum_{j=0}^n\frac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$ Now back to the coefficients, ive found it on OEIS A008955

they are called the central factorial numbers. I think this is the closest we can get for a closed form.