Example of a manifold which is not a homogeneous space of any Lie group

$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.

EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples.

Apart from already mentioned non simply connected examples most simply connected manifolds are also not homogeneous. One easy criterion is that simply connected homogeneous spaces are rationally elliptic, i.e. they have finite dimensional total rational homotopy. That is because any connected Lie group is rationally homotopy equivalent to a product of odd dimensional spheres. so a homogeneous space is elliptic by a long exact homotopy sequence.

Most simply connected manifolds are not rationally elliptic. For example the connected sum of more than two $CP^2$'s or $S^2\times S^2$'s. This is easily seen by looking at their minimal models. But even without computing minimal models it's known that an elliptic manifold $M^n$ has nonnegative Euler characteristic and has the total sum of its Betti numbers $\le 2^n$. So anything that violates either of these conditions such as the connected sum of several $S^3\times S^3$'s is definitely not rationally elliptic and hence can not be a homogeneous space or even a biquotient.

As for higher genus surfaces it should not be hard to show that they can not be homogeneous spaces $G/H$ even if you don't assume that $G$ acts by isometries. If $G/H=S^2_g$ and the $G$ action is effective then for any proper normal $K\unlhd G$ which does not act transitively on $S^2_g$ we must have $K/(K\cap H)=S^1$. But then $G/K$ is also 1-dimensional and hence also a circle which is obviously impossible. This reduces the situation to the case of $G$ being simple which can also be easily ruled out for topological reasons.

Atiyah and Hirzebruch gave a rather dramatic answer to your question in their paper "Spin Manifolds and Group Actions": if $M$ is a compact smooth spin manifold of dimension $4k$ whose $\hat{A}$-genus is nonzero then no compact Lie group can act on $M$ nontrivially, let alone transitively! The proof uses Atiyah and Bott's Lefschetz fixed point theorem in a clever way.

Unfortunately I don't have a simple example of such a manifold lying around, though I know that there are plenty of examples among 4-manifolds. It's possible that some 4 dimensional lens spaces would do the job.