That's just a long comment about tri13. I don't have Regina now, but you can get some informations by looking only at the 1-skeleton $G$ of the closed triangulation for tri13. Among the (experimentally) minimal triangulations for tri13, i.e. the ones that you find by trying to simplify it as much as you (or Regina) can, you probably find one where the 1-skeleton $G$ contains a pair of disconnecting edges, which split the $G$ into two parts $G_1$ and $G_2$. In any experimentally minimal triangulation such a pair of edges is always transverse to a torus $T$, which cut the manifold $M$ into two parts $M_1$ and $M_2$, containing $G_1$ and $G_2$ correspondingly.
According to Nathan Nathan's suggestion, the torus $T$ should separate a figure-8 knot complement (or its sibling) $M_1$ and a small Seifert space $M_2$. You can see this from the 1-skeleton $G$: the portion $G_1$ must have at least 8 vertices to be hyperbolic, and it has probably 8 or 9 (corresponding to the figure-8 knot sibling and the figure-8 knot complement), and the other portion $G_2$ has of course 5 or 4 vertices. If $G_2$ is layered, then the torus is compressible and you have a Dehn fillig of $M_1$. If (as we expect) $G_2$ is not layered, then it is certainly not a solid torus (minimal triangulations of solid tori with small number of tetrahedra must be layered, as far as I remember).
Summing up, I think that:
if a triangulation for $M$ is experimentally minimal with a reasonable number of tetrahedra and its 1-skeleton $G$ decomposes into two pieces $G_1$ and $G_2$, noone of which is layered, then you can conclude theoretically that it is not hyperbolic.
The numer of tetrahedra must be reasonable, because minimal triangulations of solid tori are classified only for small number of tetrahedra (I don't know how many...)
A fullproof fool-proof alternative is to use Matveev's Regina, which tells you exactly the JSJ decomposition of the manifold.