Here is a variant of Ian's answer. We cut along the round component in his diagram, perform a full twist (with the correct handedness), and reglue. This gives the following three-component link.

If you are seeing this in black and white, then below we call the figure-eight component "black", the round unknot "purple", and the infinity sign "green". The purple loop is a geodesic (thought of as living in the figure eight knot complement) so still bounds an embedded pair of pants $P$ with one material boundary (on purple) and two ideal boundaries (on black). The green is a geodesic immersed in $P$. So, we delete the purple loop from the link, perturb the green loop (as shown), and obtain the desired link.

Ok, checking my work in SnapPy, it seems that there is something interesting going on. Depending on how we choose the central crossing for green we get the links L8a3 or L6a1. The latter has only six tetrahedra (so that is good for the original poster's question). Also, this suggests that there is a smooth path in the space of cone manifolds that (a) connects L8a3 to L6a1 and (b) passes through the figure-eight.

Amusingly enough, this last sentence points to a much lighter proof that an example, as asked for in the original post, exists. Switching the central crossing on the green loop (and deleting purple) moves us between a pair of links with distinct complements. Performing a meridional filling on each gives us two core curves in the figure-eight knot complement. These core curves are homotopic, but not isotopic (because their complements are not homeomorphic), to each other. Thus at most one of the cores can be isotopic to a geodesic. (In fact, neither is, but that requires more work to see.).