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 $(-1,n)$.
One can also consider the asymptotic behavior of $I_{n,p,a}$ as $n\to\infty$. Writing $e^{it}-e^{-it}$ (in the expression for $g_{n,a}(t)$) as $2i\sin t$ and then writing $\sin(\frac\pi2+k\pi+h)=(-1)^k\,e^{-h^2/(2+o(1))}$ for $k=0,1,\dots$ and $h\to0$, one sees that $$I_{n,p,a}\sim\frac{C_{p,a}}{\sqrt n}\Phi(e^{ib},p+1,1/2)$$ as $n\to\infty$, where $C_{p,a}$ is manifestly nonzero and does not depend on $n$, $b$ equals $\pi a$ or $\pi(a+1)$ depending on whether $n$ is even or odd, and $$\Phi(z,s,c):=\sum_{k=0}^\infty\frac{z^k}{(k+c)^s}$$ is the Lerch transcendent; see e.g. Wikipedia. Below are pictures of the sets $\{(b,p,\Re\Phi(e^{ib},p+1,1/2)/2^p)\colon\,0.1<b<15,-0.9<p<15\}$ and $\{(b,p,\mathrm{sgn}\,\Re\Phi(e^{ib},p+1,1/2))\colon\,0.1<b<15,-0.9<p<15\}$, which suggest that $\Re\Phi(e^{ib},p+1,1/2)>0$ for all real $b\ne0$ and all real $p>-1$. (If $b=0$, then $a\in\{-1,0\}$, in which case it is easy to see that $I_{n,p,a}\ne0$.)
