This won't be possible. The smallest faithful permutation action of $Q_8$ is the regular one. That is, the smallest symmetric group containing $Q_8$ as a subgroup is $S_8$.
To see this, note if there is a subgroup $H\subseteq S_n$ isomorphic to $Q_8$, then $X=\{1,\cdots,n\}$ carries a group action from $H$. By the orbit-stabilizer theorem, if this action is transitive must be equivalent to an action on a coset space $H/K$, which is equivalent to $Q_8$ acting on $Q_8/N$ for some subgroup $N\le Q_8$. But every subgroup of $Q_8$ is normal, by inspection, so such a group actions would be unfaithful unless the kernel $N$ is trivial. If $X$ is not transitive, then it's a union of nonregular orbits, but since every proper subgroup of $Q_8$ contains the central element $-1$, we know $-1$ must be in the kernel of this action, so again it's not faithful.
Alternatively, we can find $2$-Sylow subgroups of $S_n$ and compare with $Q_8$. After all, if $S_n$ contained an isomorphic copy of $Q_8$, then it'd have to be contained in a $2$-Sylow of size at least $2^3$. For $n=4$ and hence also $n=5$ the $2$-Sylow is the dihedral group $D_8$ of order $2^3$ which is not isomorphic to $Q_8$. Indeed for $n=6$ and hence also $n=7$ the $2$-Sylow is $D_8\times C_2$ (contained in $S_4\times S_2$) which does not have a pair of noncommuting involutions generating a copy of $Q_8$.
Thus $S_8$ is the smallest symmetric group containing $Q_8$ (which is the permutation representation afforded by the left regular action, as used in Cayley's theorem).
