Is Seifert-Weber space homogeneous for a Lie group?

1. Answering your question in a comment: It is known that every compact homogenous manifold $M$ satisfies $\chi(M)\ge 0$, this was proven by Mostow, in

Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024.

2. This theorem is a consequence of a deeper structural results for homogeneous manifolds that Mostow proves in his paper. In section 4, Mostow proves that every connected homogeneous manifold (compact or not) admits an iterated fibration with fibers that are homogeneous manifolds satisfying further properties. Specializing to 3-manifolds $M$, it follows that:

i. either $M$ itself fibers, or

ii. (The exceptional case.) Either $G$ is solvable or is semisimple and in this case $H$ has finitely many connected components. Unfortunately, this exceptional case is poorly explained in the statement of the theorem in the introduction: In this case Mostow should have allowed one more subgroup in his sequence. In Mostow's notation, $k=1$, causing appearance of a subgroup $F_{-1}$ which has to be equal to $\{1\}$, while $F_0=H$. Then the fibration in the statement of Theorem C reads: $$ H/F_{-1}=H \to G/F_{-1}=G\to G/H. $$
Then $G_k=G_1=G$, $\Gamma_k=\Gamma_1=H$ and we are in Mostow's case 3, where $G=G_1$ is semisimple or solvable and, in the semisimple case, $\Gamma_k$ has finitely many components. Mostow explains this better when he repeats the formulation of Theorem C in the end of section 4 (where this theorem is actually proven).

Case i. If a compact connected 3-manifold fibers, then either the fiber is a surface and the base is a circle or vice-versa.

(a) If the fiber is a surface, it has to be of nonnegative Euler characteristic (see part 1). Hyperbolic 3-manifolds do not admit such fibrations:

Namely, asphericity of $M$ implies that the fiber $F$ cannot be $S^2, P^2$. The long exact sequences of homotopy groups of a fibration implies that we have a short exact sequence $$ 1\to \pi_1(F)\to \pi_1(M)\to \pi_1(S^1)= {\mathbb Z}\to 1. $$ If $F$ is the torus or the Klein bottle then $\pi_1(M)$ would contain a nontrivial normal virtually abelian subgroup which is known to be impossible.

(b) On the other hand, a compact hyperbolic manifold also cannot fiber over a surface $B$ with circular fibers $F$. This again follows from asphericity of $M$ which results in a short exact sequence $$ 1\to \pi_1(F)= {\mathbb Z}\to \pi_1(M)\to \pi_1(B)\to 1. $$ Then, again we would obtain a normal infinite cyclic subgroup in $\pi_1(M)$, which is impossible.

Thus, compact hyperbolic 3-manifolds cannot admit transitive actions of Lie groups.

With more work, one can relax the compactness assumption in (2) to finiteness of volume and prove a similar theorem in higher dimensions as well.

ii. Consider now the exceptional case. If $G$ is semisimple and $H$ has finitely many components, $\pi_1(M)=\pi_1(G/H)$ is finite. This is impossible for a hyperbolic manifold. Lastly, if $G$ is solvable, so is $H$, hence, $H/H^c$ is solvable too (I am using Mostow's notation where $H^c$ denotes the identity component with respect to the Lie group topology). But then $\pi_1(M)$ is solvable, which is impossible for a hyperbolic manifold of finite volume.