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Surely, this number is at least $\frac{n^2+n+2}{2}$, because a generic collection of lines cut the plane in this amounts of pieces. This is a bit more than Harnak's bound and I would be surprised if it can be beaten.

One can prove a quadratic bound on the number of regions (quite possibly close to $\frac{n^2+n+2}{2}$) by the following argument:
We will assume for simplicity that $X$ is in $\mathbb RP^2$. Take a generic point $p$ on $\mathbb RP^2$ and consider a pencil of lines through $p$. Consider the projection of $X$ to $\mathbb RP^1$. Notice that the number of components of $\mathbb RP^2\setminus X$ is related to the number of critical points of this projection. (If $x\in X$ is a singular point of $X$ then the $x$ is also considered to be critical). Indeed if there where no critical points at all, the number of components of $\mathbb RP^2\setminus X$ would be at most $n$. So the number of components is proportional to the number of critical points of the projection. The last number is at most quadratic.
Surely, this number is at least $\frac{n^2+n+2}{2}$, because a generic collection of lines cut the plane in this amounts of pieces. This is a bit more than Harnak's bound and I would be surprised if it can be beaten.