This doesn't answer your question completely, but at least it's a start.
If $G$ is a complex Lie group, then its quotient $G/H$ by a complex subgroup $H$ is always a complex manifold. In case $G$ is semisimple (and connected), then the compact quotients $G/H$ come from what are called parabolic subgroups $H$. For simplicity, let's assume that $G$ is simple. Then, up to conjugacy, the parabolic subgroups of $G$ lie in one-to-one correspondence with subsets of the nodes of the Dynkin diagram of $\mathfrak{g}$. It's easy to find a description of this bijection in the literature, so I will say no more about it. Let me just mention that a parabolic subgroup corresponding to omitting one node from the Dynkin diagram is called maximal parabolic.
Now suppose that $K$ is a compact real form of the complex semisimple group $G$. Then if the quotient space $G/H$ is compact (i.e., if $H$ is parabolic), a theorem of Montgomery asserts that $K$ acts transitively on $G/H$, and so we obtain an identification $G/H = K/(K\cap H)$. The a priori real manifold $K/(K\cap H)$ is thus endowed with a complex structure.
In summary, we have the following result:
Let $K$ be a compact semisimple Lie group with complexification $G$. Then, for a closed subgroup $S$ of $K$, the homogeneous space $K/S$ is complex if $S = K \cap P$ for some parabolic subgroup $P$ of $G$.
Here's an example. Let $G = \rm{SL}(n,\mathbb{C})$. In terms of the bijection between parabolic subgroups of $G$ and nodes, it's not hard to see that the maximal parabolic corresponding to deleting the $k$th node is the subgroup of $G$ preserving a $k$-dimensional subspace of $\mathbb{C}^n$. In particular, if we let $P$ denote the maximal parabolic corresponding to the deletion of the first node, we get $G/P = \mathbb{CP}^{n-1}$. Now the maximal compact subgroup of $G$ is $K = \rm{SU}(n)$. In an appropriate basis, $P$ is the subgroup of $G$ of the form
$$ \begin{pmatrix} \ast & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \\ \vdots & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \end{pmatrix}. $$
Thus $K \cap P = \rm{U}(n-1)$ (to see this quickly, just recall that unitary matrices leave invariant the perp of an invariant subspace), where I'm thinking of $\rm{U}(n-1)$ as sitting in $K$ as
$$ \begin{pmatrix} (\det A)^{-1} & 0 \\ 0 & A \end{pmatrix}. $$
So by the result stated above we obtain the identification $\rm{SU}(n)/\rm{U}(n-1) = \mathbb{CP}^{n-1}$ given in the OP. Using the other maximal parabolics we can get a similar description for the complex Grassmannian of $k$-planes in $\mathbb{C}^n$.
Here are a couple of useful references:
Wang, Hsien-Chung, Closed manifolds with homogeneous complex structure. Amer. J. Math. 76 (1954). 1–32.
Wolf, Joseph A, The action of a real semisimple group on a complex flag manifold. I. Orbit structure and holomorphic arc components. Bull. Amer. Math. Soc. 75 (1969), 1121–1237.
Both these papers address the issue of finding complex structures on compact homogeneous spaces (so their results apply in particular to homogeneous spaces coming from compact Lie groups). There's a standing assumption of simply connectedness in Wang, but if I remember correctly, there is no such restriction in Wolf's paper.
Edit: Here's some more information that might be helpful. Let $K$ and $G$ be as above. A parabolic subgroup $P$ of $G$ admits "Levi decompositions" $P=LU$, where $U$ is the unipotent radical of $P$ and $L$ is a reductive group (called a Levi factor). In particular, every Levi factor $L$ has a compact real form $L_{\mathbb{R}}$. The different Levi factors of $P$ are all conjugate. There is a choice that is compatible with the choice of maximal compact $K$, and in this case we have $P \cap K = L_{\mathbb{R}}$.
