You have to be careful with what you mean here. As your equations involve
complex conjugation they do not define a *complex* variety. They do
define a real algebraic variety. However, then you have to be careful because
there may be reducible components that "are not seen" easily in the real
picture, they may not have real points, only complex ones (to be precise all
their real points may lie on othere irreducible components). This is what
happens in your example.

The first step in understanding a real algebraic variety is usually to extend
scalars to the complex numbers. To understand what happens in your case it pays
to describe a little bit more abstractly. Hence considered over the reals what
we have is a finite dimensional $\mathbb R$-algebra with an $\mathbb R$-involution, namely
$\mathbb C$. We then consider $n\times n$-matrices over $\mathbb C$ which inherits an
involution from the involution of $\mathbb C$, $M\mapsto M^\dagger$. As it is
$\mathbb R$-linear the condition $MM^\dagger=M^\dagger M$ is given by
polynomials with real coefficients and hence defines a real algebraic subvariety
of $M_n(\mathbb C)$ which is seen as the affine space
$\mathbb A^{2n^2}_{\mathbb R}$ over the reals. When we extend scalars we get exactly the same description
but replacing the involutive $\mathbb R$-algebra $\mathbb C$ with its scalar
extension $\mathbb C\bigotimes_{\mathbb R}\mathbb C$. This algebra is isomorphic
to $\mathbb C\times\mathbb C$ with the involution permuting the two
factors. Hence $M$ is now described by two complex matrices $(A,B^t)$ and
$(A,B^t)^\dagger=(B,A^t)$ and $MM^\dagger=M^\dagger M$ is turned into
$(AB,B^tA^t)=(BA,A^tB^t)$ which just describes pairs $(A,B)$ of commuting
matrices. Hence the complex scalar extension is the variety of pairs of
commuting matrices with a non-standard real structure where complex conjugation
takes $(A,B)$ to $(\overline B,\overline A^t)$ (instead of $(A,B)\to (\overline
A,\overline B)$ which is the more standard one).

Now, it is well-known that the variety of pairs of commuting matrices is
*very* complicated. There is a more easily understood subset consisting
of the pairs $(A,B)$ where both $A$ and $B$ are semi-simple. From an algebraic
point of view it is not as it a subvariety (it is a constructible but not
locally closed subset). It does contain the open algebraic subset of pairs
$(A,B)$ where all the roots of the characteristic polynomials of $A$ (say) are
distinct. That subset is Zariski dense in the set of all semi-simple matrices
and hence the latter lie in the closure of the former. All the real points will
then lie in this closure $\overline S$. However, it is well-known that unless
$n$ is *very* small the closure is not equal to the whole variety (there
are pairs that can not be deformed into semi-simple ones). This means that while
$\overline S$ is an irreducible component (that is defined over the real
numbers) there are (many) other irreducible components. They will be exactly of
the type described above, all their real points will also lie in $\overline S$.

**Addendum**: As to the question of dimension, of course the closure of semi-simple matrices has the right dimension but I think (but don't quite remember) that there are other components of larger dimension.