We can achieve 6 by projecting onto a plane perpendicular to an edge of the dodecaheron. Indeed, if we imagine viewing the dodecahedron in such a direction, then 4 of the faces are projected to line segments (namely, the two faces adjacent to the edge and the two opposite faces), and of the remaining 8 faces, 4 appear on the front of the dodecahedron and the other 4 are on the back. Thus, the dodecahedron appears as shown. To see that we cannot do better, note that, by central symmetry, the number of edges of the projection must be even. So we just need to show that the answer cannot be 4. But if the projection had 4 sides, one of the vertices would give a projection forming an acute angle, which is not possible. So 6 is the answer.