Smallest graph with automorphism group the quaternion $8$-group, $Q_8$
Edit: 16 vertices is indeed the smallest possible size for such a graph (see the proof below).
Here's an example of a graph with 16 vertices whose automorphism group is $Q_8$. As we will show, the automorphisms of the graph preserve the colors and arrows:
The symmetry of this graph is similar to the symmetry of the Cayley graph of $Q_8$ shown in the question above. Here's an argument that it's possible to recover the colors and arrows from the structure of the graph itself:
Cyan vertices have degree 7, while yellow vertices have degree 5.
Black edges connect pairs of yellow vertices.
Green edges connect vertices of different colors and do not share triangles with black edges.
Purple edges connect cyan vertices and are involved in triangles with green edges.
Red edges connect vertices of different colors and are involved in triangles with green and purple edges.
Blue edges connect vertices of different colors and are neither red nor green.
Cyan edges connect cyan vertices and are not purple.
Each black edge is involved in exactly one black-red-blue triangle, and the arrow points from the vertex opposite the red edge to the vertex opposite the blue edge.
Using the colors and arrows, it's easy to prove that any automorphism of this graph that fixes a vertex must fix all 16 vertices, so there are at most eight automorphisms. It's easy to check that there are in fact eight automorphisms, which form a copy of $Q_8$.
Incidentally, Babai constructs a 16-vertex graph for $Q_8$ in his paper "On the minimum order of graphs with given group" with 52 edges. The graph above is a slight improvement on this, having the same number of vertices but only 48 edges. Though 16 is the minimum possible number of vertices, it would be interesting to know whether there's a 16-vertex graph with fewer than 48 edges that has symmetry group $Q_8$.
Edit: 16 vertices is indeed the minimum possible. Indeed, any graph whose automorphism group is $Q_8$ must have at least two orbits of size $8$.
To see this, observe first that any faithful action of $Q_8$ on a graph must have at least one orbit of size $8$, since otherwise the action factors through $Q/\{\pm 1\}$.
Now, suppose we have a faithful action of $Q_8$ on a graph $G$, and suppose that $G$ has just one orbit of size $8$. Let $v$ be a vertex in this orbit. I claim that switching $v$ with $-1v$ is an automorphism of $G$. This automorphism is not in $Q_8$, since the action of $Q_8$ on the orbit of size $8$ is the left regular action, so it follows that the automorphism group of $G$ is larger than $Q_8$.
To prove that switching $v$ and $-1v$ is an automorphism of $G$, observe that:
If $w$ is any vertex not in the orbit of size $8$, then the stabilizer of $w$ is a proper subgroup of $Q_8$, so $-1w = w$. Thus, there is an edge from $v$ to $w$ if and only if there is an edge from $-1v$ to $w$.
If there is an edge from $v$ to $iv$, then its image under $i$ is an edge from $iv$ to $-1v$. Thus $v$ is connected by an edge to $iv$ if and only if $-1v$ is connected by an edge to $iv$. The same holds for $-iv$, $\pm j v$, and $\pm k v$.
This proves the claim, and therefore 16 is the smallest possible size.
What is the minimal number of vertices of a graph $\Gamma$ for which $\text{Aut}(\Gamma) \cong Q_8$? What is an example of such a graph that achieves this minimum? Among such graphs, which has (have) the minimal number of edges?
There has been progress on this question since I asked it three years ago: While Jim Belk's answer establishes that a graph $\Gamma = (V, E)$ with automorphism group $Q_8$ must have $|V| \geq 16$ (I don't know the earliest reference for this fact), Graves, Graves, & Lauderdale recently showed (2018) that the graphs with $|V| = 16$ and $\operatorname{Aut}(\Gamma) \cong Q_8$ satisfy the sharp bound $|E| \geq 44$.
Probably the article includes an example of a graph realizing the above minimum, but I cannot find an ungated copy and am away from my institution at the moment.
This REU presentation shows that if we allow more vertices we can find a graph with $\operatorname{Aut}(\Gamma) \cong Q_8$ and $|E| = 40$ and that this is a sharp lower bound for graphs with $|V| \leq 20$.
Graves, C.; Graves, S.J.; Lauderdale, L-K. "Smallest graphs with given generalized quaternion automorphism group." J Graph Theory. 87 (2018), 430-442. https://doi.org/10.1002/jgt.22166