This is only a very partial answer. Suppose that $a<-n$ and $p$ is a non-integer in the interval $(-1,n)$. Then $2j-n-a>0$ for all $j=0,\dots,n$. So, by the mean-value theorem applied (say, repeatedly) to the $n$-fold symmetric difference in the expression of $I_{n,p,a}$ in $(**)$ in the question statement, one has $$I_{n,p,a}=i^p\, \Gamma (-p)\, p(p-1)\cdots(p-n+1)2^n(2j_{n,p,a}-n-a)^{p-n}$$ for some real $j_{n,p,a}\in(0,n)$, so that $$i^{-p}(-1)^n I_{n,p,a}=\Gamma (n-p)\,2^n(2j_{n,p,a}-n-a)^{p-n}>0.$$ Quite similarly, $$i^p(-1)^n I_{n,p,a}=\Gamma (n-p)\,2^n(n+a-2j_{n,p,a})^{p-n}>0$$ if $a>n$ and $p$ is a non-integer in the interval $(-1,n)$. So, $I_{n,p,a}\ne0$ if $|a|>n$ and $p$ is a non-integer in the interval $(-1,n)$.
By using the l'Hospital rule as mentioned in the question statement, one should likely get the same result when $|a|>n$ and $p$ is an integer in the interval $(-1,n)$.