Suppose that $\varpi=\frac{1}{2}(p-1).$ By $$(p-1)!=1\cdot2\cdots\frac{1}{2}(p-1)\{p-\frac{1}{2}(p-1)\}\{p-\frac{1}{2}(p-3)\}\cdots(p-1)\equiv(-1)^\varpi(\varpi!)^2(\bmod p),$$ it follows, by Wilson's theorem, that $$(\varpi)^2\equiv(-1)^{\varpi-1}(\bmod p).$$ Now we need to distinguish the two cases $p=4n+1$ and $p=4n+3$.
If $p=4n+3$ as in your question, then $$(\varpi!)^2\equiv1(\bmod p),$$ $$\varpi!\equiv\pm1(\bmod p).$$ Since $-1$ is anon-residue of $p$, thr sign in the above conruence ispositive or negative according as $\varpi!$ is a residue or non-residue of $p$.
As we knew it, a product of two residues or non-residues of $p$ is also a residue, while the product of one residue and a non-residue of $p$ is no longer a residue. Therefore, the sign is positive or negative according as the number of non-residues of $p$ less than $\frac{1}{2}p$ is even or odd.
The case $p=4n+1$ remains nearly the same argument.