Very short answer: You have to be very precise about which base field, $\mathbb R$ or $\mathbb C$, you are considering in each case.
Over $\mathbb C$, there is the Lie group $SL_n(\mathbb C)$ and its Lie algebra $\mathfrak{sl}_n(\mathbb C)$, and every Cartan subalgebra of this will have roots which form a system of type $A_{n-1}$. There is extensive literature on this.
Over $\mathbb R$ however, one can e.g. look at the Lie groups $SL_n(\mathbb R)$, which have Lie algebra $\mathfrak{sl}_n(\mathbb R)$, but also at the Lie groups $SU(n)$ and their Lie algebras $\mathfrak{su}_n$ -- note that elements of these are often written as certain matrices with complex entries, but they are not complex Lie groups resp. algebras, but real ones. Notice in particular that $\mathfrak{su}_n$, which indeed can be identified with the traceless skew-hermitian $n\times n$-matrices, is not a vector space over $\mathbb C$, but over $\mathbb R$ (of dimension $n^2-1$).
Now it turns out that the non-isomorphic real Lie groups $SL_n(\mathbb R)$ and $SU_n$ both have complexification (isomorphic to) $SL_n(\mathbb C)$. They are so-called real forms of $SL_n(\mathbb C)$. Likewise but even simpler to see -- on the Lie algebra level, complexification is just done by tensoring with $\mathbb C$ -- both $\mathbb C \otimes_{\mathbb R}\mathfrak{su}_n$ and $\mathbb C \otimes_{\mathbb R}\mathfrak{sl}_n(\mathbb R)$ are isomorphic to $\mathfrak{sl}_n(\mathbb C)$, i.e. both $\mathfrak{su}_n$ and $\mathfrak{sl}_n(\mathbb R)$ are real forms of $\mathfrak{sl}_n(\mathbb C)$.
For $n=2$, $SL_2(\mathbb R)$ and $SU_2$ are (up to isomorphism) the only real forms of $SL_2(\mathbb C)$. For higher $n$ though, and for other classes of Lie groups / algebras, there are usually more real forms. The last example here is a real form of $\mathfrak{sl}_3(\mathbb C)$, called $\mathfrak{su}_{1,2}$, which is neither isomorphic to $\mathfrak{sl}_3(\mathbb R)$ nor to $\mathfrak{su}_3$.
It's quite common in the literature when speaking of root systems, what is meant is actually the root system of the complexification. In that terminology, both $SL_n(\mathbb R)$ and $SU_n$ (or their Lie algebras) have root system $A_{n-1}$. However, there is also the notion of relative or restricted or real or $k$-rational (here for $k=\mathbb R$) root systems; in this case, the relative root system of $SL_n(\mathbb R)$ would still be $A_{n-1}$, whereas the relative root system of $SU_n$ is empty (which is always the case for compact semisimple groups). More on those "relative roots" e.g. here, where I tried to compute all examples of real forms where that restricted root system is of type $BC$ (something that can never happen for complex Lie groups / algebras).
One further thing to note: By a fantastic coincidence (?), for each complex simple Lie algebra, there is up to iso exactly one real form which is compact (e.g. above, $\mathfrak{su}_n$ is the compact real form of $\mathfrak{sl}_n(\mathbb C)$). Also, there is always exactly one so-called "split" real form, whose restricted roots are just the same as the roots of the complexified version (e.g. above $\mathfrak{sl}_n(\mathbb R)$ is the split real form of $\mathfrak{sl}_n(\mathbb C)$). In a way, these two are extreme cases on opposite ends of a spectrum. As noted above, in general there are many more cases "between" them. They are classified by so-called "Satake diagrams", which are like an upgrade of the Dynkin diagrams: the underlying Dynkin diagram of a Satake diagram tells us of what type ($A_n, B_n, C_n, ..., G_2$) the complexification is, and the extra ornaments that make it a Satake diagram (black vs. white nodes, and arrows) encode which real form of that complex type we have. See further references and examples here or here.
Added: It's maybe worthwhile to note that beyond everything mentioned above, the (Lie group / Lie algebra)-correspondence is also not one-to-one, over any ground field. Rather, for one given semisimple Lie algebra there is a lattice of connected groups that "sits" over it, with an adjoint (centreless) one at the bottom and a simply connected one on top. E.g. over $\mathbb C$,
$PSL_2(\mathbb C)$ (adjoint) and $SL_2(\mathbb C)$ (simply connected) share the Lie algebra $\mathfrak{sl}_2(\mathbb C)$;
whereas over $\mathbb R$,
$PSL_2(\mathbb R)$ (adjoint), $SL_2(\mathbb R)$, $Mp_2(\mathbb R)$ (the metaplectic group), ... , $\overline {SL_2(\mathbb R)}$ (the simply connected universal cover of $SL_2(\mathbb R)$), with the "..." being infinitely more in-between, all share the Lie algebra $\mathfrak{sl}_2(\mathbb R)$ (compare last three sentences here);
whereas the compact real one has only two manifestations again:
$PSU_2$ (adjoint, and happens to be $\simeq SO_3(\mathbb R)$) and $SU_2$ (simply connected) share the Lie algebra $\mathfrak{su}_2$.
If one allows even disconnected groups, then there's infinitely many more groups sitting over each Lie algebra, but that's basically stuff like
$SL_2(\mathbb C) \times$ (your favourite finite group) still has Lie algebra $\mathfrak{sl}_2(\mathbb C)$.
Best Answer
s.harp is right that the space is diffeomorphic to $\coprod_{k=0}^n U(n)/(U(k)\times U(n-k))$. Call $X:=U(n)\cap\mathfrak{u}(n)$.
Proof: Consider the $U(n)$ action on $C^{n^2} = M_n(\mathbb{C})$ by conjugation. This action preserves $U(n)\subseteq M_n(\mathbb{C})$ becuase $U(n)$ is a subgroup. It also preserves $\mathfrak{u}(n)\subseteq M_n(\mathbb{C})$ because this is nothing but the adjoint action.
It follows that the $U(n)$ conjugation action maps $X$ to itself. Further, since every element in $U(n)$ is conjugate to a diagonal element, every orbit contains a diagonal matrix. Further, each of the permutation matrices is in $U(n)$, so we may further conjugate the diagonal matrix to put the diagonal elements in any order we like.
As noted in the comments, the eigenvalues must be $\pm i$. (One way to see this: the eigenvalues of an element of $U(n)$ are all norm $1$, while the eigenvalues of an element in $\mathfrak{u}(n)$ are purely imaginary.) Since conjugation preserves the eigenvalues, and because every element of $U(n)$ is conjugate to a diagonal matrix we have proven:
(So far, all of this was essentially in the comments.)
Now, by compactness, the $U(n)$ orbits through distinct $E_k$s have a positive distance between them. So $X$ is given as a disjoint union of orbits through the $E_k$.
Finally, we use this wonderful fact that for a compact Lie group action, the orbit $U(n)\cdot E_k$ is canonically diffeomorphic to $U(n)/U(n)_{E_k}$, where $U(n)_{E_k}$ denotes the stabilizer at $E_k$. That is, $U(n)_{E_k} = \{B\in U(n): BE_k B^{-1} = E_k$.}
So, we need only show that $U(n)_{E_k} = U(k)\times U(n-k)$. Given $B\in U(n)$, write $B$ in the block form $B = \begin{bmatrix}C & D\\ E & F\end{bmatrix}$ with $C$ a $k\times k$ matrix and $F$ an $(n-k)\times (n-k)$ matrix.
Then \begin{align*} B\in U(n)_{E_k}&\iff BE_k = E_k B\\ &\iff iD = -iD\text{ and } -iE = iE\\ &\iff D = 0\text{ and } E = 0\\ &\iff B\in U(k)\times U(n-k).\end{align*}
$\square$
The spaces $U(n)/(U(k)\times U(n-k))$ are called complex Grassmannians and have the geometric interpretation of being the set of all $k$-dimensional subspaces in $\mathbb{C}^n$. For certain values of $n$ and $k$, they have other descriptions.
For example, for any $n$, $U(n)/U(0)\times U(n)$ and $U(n)/U(n)\times U(0)$ are both points (as you saw in the $n=1$ and $n=2$ case). Further, for any $n$, $U(n)/U(1)\times U(n-1)$ and $U(n)/U(n-1)\times U(1)$ are both diffeomorphic to $\mathbb{C}P^{n-1}$. Note that when $n=2$, $\mathbb{C}P^1 = S^2$, as you saw.