This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.
Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)
Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.
[EDIT] (The above is almost true except there are other components.) Consider first the case when $Y$ and $Z$ are general. We may chose a square root $B^{1/2}$ and consider matrices $$x=B^{1/2}XB^{1/2},\,\,a=B^{1/2}AB^{1/2}.$$ In this terms, the above equation is $$xa=ax,\,x^2+ax+a=0.$$ In the general case, it has $2^n$ solutions (consider $a$ in the diagonal form.) I presumed that all these solutions are Galois conjugate, but on second thought, no. At least, there are two obvious solutions $X=-Y$ and $X=-Z$ which are not conjugate to anything. For $n>1$ there are more solutions.
So, there are four components, $X+Y=0, X+Z=0, Y+Z=0$ and the fourth one (probably reducible) comes from extra solutions of the above equation (for $n>1$). So, it is not much. Still, the above argument has a point: it should be possible to count the components once you figure out how to divide these $2^n$ solutions between them.