We claim that every pair of elements of $A$ must belong to $C$, so that the answer is $\binom{n}{2}$. Indeed, if $a, b \in A$ and $\{a, b\}$ is not in the 2-configuration, then we can assign the other elements of $A$ the numbers $1,2, \ldots, n-2$ and assign $a$ and $b$ both the number $n-1$, so that $C$ is $(n-1)$-separable. On the other hand, if every pair of elements of $A$ is in the configuration, then $A$ cannot be $m$-separable for $m<n$, since this would require assigning the same number to at least two elements, and then we would have a pair whose elements have the same number.