Because you're assuming the group is compact, there's another method for classifying such spaces. This approach can be used in higher dimensions (though people often assume the resulting manifold is simply connected just to cut down on the work).
I'll also always be assuming things are connected: the components of a homogeneous space $G/H$ are always diffeomorphic, and the identity component of $G$ acts transitively on a connected component of $G/H$, so this is all ok.
Suppose $G$ is a connected compact Lie group, $H\subseteq G$ is a closed subgroup. Because $G$ is compact, we may equip the homogeneous space $G/H$ with a Riemannian metric for which the action by $G$ is isometric.
Now, consider the isotropy action of $H$ on $T_{eG} G/H$, where $e\in G$ denotes the identity element. Recall (see, e.g., this MSE question) that if an isometry of a (connected) Riemannian manifold fixes a point and acts as the identity at the tangent space at that point, then that isometry must be the identity. Elements of $h$ fix $e H\in G/H$, so, if such an $h$ acts trivially on $T_{eG}G/H$, we must have $h = e$.
Said another way, the isotropy action gives an injective map $H\rightarrow O(T_{eG} G/H)$ into the orthogonal group. Thus, we may view $H$ as a subgroup of $O(T_{eG} G/H)$.
I'll now specialize to the case where $G/H$ is $3$-dimensional, so $O(T_{eG} G/H)$ can be identified with $O(3)$. The connected subgroups of $SO(3)$ are well known: they are $\{e\}, SO(3)$, or a conjugate of the usual $SO(2)\subseteq SO(3)$. Note that $G/H$ is equivariantly diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$, so, in terms of classification, we can assume the identity component $H^0$ of $H$ is given by one of $\{e\}, SO(2)$, or $SO(3)$.
We'll break into cases depending on $H^0$.
Case A
If $H^0 = \{e\}$, then $3 = \dim G - \dim H$ impies that $\dim G = 3$. From the classification of simply connected Lie groups, $G$ has a cover which is either isomorphic to $SU(2)$ or to $T^3$. In the first case $G = SU(2)$ or $SO(3)$ and $SO(3)/H = SU(2)/\pi^{-1}(H)$ with $\pi:SU(2)\rightarrow SO(3)$ the double cover, so you just get quotients of $SU(2)$. In the second case, you get quotients of $T^3$. However, $H\subseteq T^3$ is normal, so $T^3/H$ is an abelian Lie group of dimension $3$, so $T^3/H\cong T^3$ no matter what $H$ is (so long as $H^0 = \{e\}$.)
Case B
If $H^0 = SO(2)$, then $\dim G = 4$. From the classification of simply connected Lie groups, $G$ has a cover of the form $G = T^4$ or $G = SU(2)\times S^1$. If $G = T^4$, the argument from the previous paragraph establishes that $T^4/H\cong T^3$. So, we will assume that $G$ is covered by $G = SU(2)\times S^1$. In fact, we'll pull everything back: $G/H\cong (SU(2)\times S^1)/\pi^{-1}(H)$, so we'll actually work with $G= SU(2)times S^1$. (Note though, that by pulling back, $\pi^{-1}(H)$ need not act effectively on $(SU(2)\times S^1)/\pi^{-1}(H)$) any more.) Also, instead of writing $\pi^{-1}(H)$ everywhere, I'll abuse notation and just write $H$.
Lemma: Suppose $H$ acts diagonally on $G_1\times G_2$ and that the action of $H$ on $G_2$ is transitive with isotropy group $H'$. Then $(G_1\times G_2)/H$ is canonically diffeomorphic to $G_1/H'$.
Proof: Just check that the map $G_1/H'\rightarrow (G_1\times G_2)/H$ sending $g_1 H'$ to $(g_1,e)H$ is a diffeomorphism. $\square$
Using the lemma, it follows that if the projection of $H^0$ to the $S^1$ factor of $G$ is surjective, then we can write $G/H\cong SU(2)/H'$, where $H'$ is $0$-dimensional. These were classified in Case A). So, we may assume that the projection of $H^0$ to the $S^1$ factor of $G$ is trivial. That is, we may assume $H^0$ acts only on the $SU(2)$ factor of $G$, so $G/H^0\cong S^2\times S^1$.
So, we understand $H^0$ in this case, but what about $H$? Well, let's assume $H\neq H^0$, and pick $(h_1,h_2)\in H\setminus H_0$. Because the normalizer $N:=N_{SU(2)}(SO(2))$ has two components, and $h_1 \in N$, we know that $(h_1,h_2)^2$ acts as a rotation on just the $S^1$ factor of $G/H^0 \cong S^2\times S^1$. Hence, $(G/H^0)/\langle (h_1,h_2)^2\rangle \cong S^2\times S^1$, so we may as well assume that $(h_1,h_2)^2 \in H^0$, so that, in particular, $h_2 = \pm 1$. That is, we may as well assume that $H$ has precisely two components. If $h_1\in SO(2)$, then $G/H$ is diffeomorphic to $S^2\times S^1$ independent of $h_2$. If $h_1\notin SO(2)$ and $h_2 = 1$, we get $\mathbb{R}P^2\times S^1$, and if $h_1\notin SO(2)$ and $h_2 = -1$, we get the non-trivial $S^2$-bundle over $S^1$ (the quotient of $S^2\times S^1$ by the diagonal antipodal action).
This concludes case B.
Case C In this case, $H^0 = SO(3)$. Then $\dim G = 6$, so $G$ is covered by one of $T^6, SU(2)\times T^3$, or $SU(2)\times SU(2)$.
For $G=T^6$, because $H^0$ is non-abelian, there is no $H^0\subseteq G$, even up to cover.
For the other two, by replacing $H^0$ with a cover of $H^0$ (i.e., $SU(2)$), we can actually assume $G= SU(2)\times T^3$ or $G = SU(2)\times SU(2)$. However, in both cases, it is easy to see that the above Lemma applied, so we find $G/H\cong T^3/H'$ or $SU(2)/H'$ where $H'$ is $0$-dimensional, getting us back to Case A. This completes Case C.
Best Answer
The answer is yes.
In fact something much stronger is true. A compact smooth homogeneous $ K(\pi,1) $ with fundamental group $ \pi $ nilpotent is determined up to diffeomorphism by $ \pi $. See Corollary 2.7 and the discussion above Theorem 2.1 in [Gorbatsevich, ON LIE GROUPS, TRANSITIVE ON COMPACT SOLVMANIFOLDS].
So, in particular, a compact smooth homogeneous $ K(\mathbb{Z}^n,1) $ must be diffeomorphic to the torus $ T^n $ with the standard smooth structure.
The intuition here is that if $ G $ acts transitively on a compact $ K(\pi,1) $ and $ \pi $ is solvable then we expect that the maximal solvable subgroup of $ G $ will also act transitively (and it is a theorem of Auslander that compact solvmanifolds are determined up to diffeomorphism by $ \pi $). Compare this to theorem of Montgomery that if $ G $ acts transitively on a compact manifold $ M $ with $ \pi_1(M) $ compact (i.e. finite since the fundamental group is discrete) then the maximal compact subgroup of $ G $ also acts transitively. Anyway the idea here is just that knowing what kind of Lie groups can act transitively on your manifold often tells you cool stuff about the topology of the manifold.
So that's the intuition but the details are a bit complicated.
Gorbatsevich conjectures that any compact smooth homogeneous $ K(\pi,1) $ with $ \pi $ solvable is determined up to diffeomorphism by $ \pi $. The idea is to prove that such a space admits a transitive action by a solvable Lie group and then use the result of Auslander that a compact solvmanifold is determined up to diffeomorphism by its fundamental group. But he says that he can't figure out how to make the proof work :(
He does, however, prove that a compact smooth homogeneous $ K(\pi,1) $ with $ \pi $ solvable is determined up to homeomorphism by $ \pi $ [Proposition 1.5].
For the case of diffeomorphism, which is what concerns us here, he does the proof for the weaker case of $ \pi $ virtually nilpotent (which is good enough for us because a torus even has $ \pi $ abelian!). He notes that every hypersolvable group is virtually nilpotent so this covers some important solvable non-nilpotent cases anyway.