 Solution

Every nontrivial subgroup of the quaternion group contains $-1$. Indeed, it either contains $-1$ directly, or contains one of $\pm i,\pm j,\pm k$, whose square is $-1$.

Suppose $Q_8$ acts on an $n$-point set with $n<8$. Every group orbit has size at most $n<8$. By the orbit-stabilizer theorem, each point's stabilizer subgroup has order $8/|Q_{8}x|>1$, so it contains $-1$. Consequently $-1$ fixes every point, and the group action is not faithful. An embedding in $S_n$ would give a faithful group action on $n$ letters, which is impossible. ThusThe left-regular group action on eight points is faithful by Cayley theorem, so the minimal faithful permutation degree of the quaternion group is exactly eight. The proof allows multiple group orbits; restricting only to transitive actions would not suffice.