= 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 $8/|Q_8x|>1$, 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. Thus
$$
\boxed{Q_8\not\hookrightarrow S_n\quad(n<8).}
$$
The 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.
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