Solution:
This is obtainable by construction. Consider two tetrahedrons glued along a face; this gives us 5 vertices, 9 edges, and 6 faces, for a total of 20, and one readily checks that the required Eulerian path exists.

Now, to see that we cannot do better, first notice that the number $v$ of vertices is at least 5, since otherwise we must have a tetrahedron, which does not have an Eulerian path. Each vertex is incident to at least 3 edges, and in fact, since there is an Eulerian path, all except possibly two vertices are incident to an even number of edges. So the number of edges is at least $(3+3+4+4+4)/2$ (since each edge meets two vertices) $=9$.

Finally, if $f=4$ then each face must be a triangle, because there are only 3 other faces for it to share edges with, and we are again in the case of a tetrahedron, which is impossible; therefore $f \geq 5$. So $f+v+e \geq 5+5+9=19$. But since $f+v-e=2-2g$ (where $g$ is the number of holes in the polyhedron), $f+v+e$ must be even. This strengthens our bound to 20 as needed.