Answer: 2.
Let's consider the subset $S = \{(p+1)/2, \dots, p-2, p-1\}$. Then
$$\Pi_s = (-1)^{\frac{p-1}{2}} \left(\frac{p-1}{2}\right)! = - \left(\frac{p-1}{2}\right)! \equiv a \pmod{p},$$
$$\overline{\Pi}_s = \left(\frac{p-1}{2}\right)! \equiv -a \pmod{p}.$$
By the Wilson's theorem $a^2 \equiv 1 \pmod{p}$. Hence, two cases are possible:
Case 1. $a \equiv 1 \pmod{p}$. Then the deviation of the essential subset $S$ is equal to $\Delta_s = 2 \pmod{p}$.
Case 2. $a \equiv -1 \pmod{p}$. Then the deviation of the essential subset $T$, where $T$ is obtained from $S$ by substituting $\frac{p+1}{2}$ to $\frac{p-1}{2} \equiv -\frac{p+1}{2} \pmod{p}$, is equal to $\Delta_T = \Pi_T - \overline{\Pi}_T \equiv 1 - (-1) \equiv 2 \pmod{p}$.
Now we prove that $2$ is the least possible remainder.
Note that $\Pi_s \cdot \overline{\Pi}_s = -1$ by the Wilson's theorem again. Since $-1$ is not a quadratic residue by modulo a prime number of the kind $4k+3$, we have certainly $\Pi_s \neq \overline{\Pi}_s \pmod{p}$. In other words, the deviation is not divisible by $p$. Suppose to the contrary, let $S$ be a subset with deviation $\Pi_s - \overline{\Pi}_s = 1$. Then $\Pi_s \equiv m+1 \pmod{p}$, $\overline{\Pi}_s \equiv m \pmod{p}$ for some $m \in \{2, 3, \dots, p-3\}$ and $m^2 + m \equiv -1 \pmod{p}$. This implies $m^3 \equiv 1 \pmod{p}$ and by the Fermat's (little) theorem $p-1$ is divisible by $3$, which is a contradiction.