I just stumbled upon this old question, and thought I would add a simple and natural example, which is $G_{\delta \sigma}$ but neither $G_{\delta}$ nor $F_{\sigma}$.

Consider $f\colon \mathbb{C}\to\mathbb{C}; z\mapsto e^z$, and its iterates
$$f^n(z) = \underbrace{f(f(\dots f(z)\dots))}_{\text{$n$ times}}=e^{e^{\cdot^{\cdot^{\cdot^{e^z}}}}}.$$
Consider the set
$$ X := \{z\in\mathbb{C}\colon f^n(z)\not\to\infty\} $$
and its subset
$$ X_0 := \{z\in\mathbb{C}\colon \{f^n(z)\colon n\geq 0\} \text{ is dense in $\mathbb{C}$}\}.$$

It follows from the definitions that $X_0$ is a $G_{\delta}$ and $X$ is a $G_{\delta \sigma}$. It is well-known that $X_0$ is dense in $\mathbb{C}$, as is the complement $I(f) = \mathbb{C}\setminus X$. (For an elementary proof, see *The exponential map is chaotic*, Amer. Math. Monthly 2017, arxiv:1408.1129.)

So $X_0$ contains a dense $G_{\delta}$, and can therefore not be $F_{\sigma}$, as noted by Pietro Majer.

That $X$ is not $G_{\delta}$ follows from the fact that $I(f)$ is not $F_{\sigma}$. (*Escaping sets are not sigma-compact*, arxiv:2006.16946.)

(More generally, for any transcendental entire function, the set of non-escaping points is a $G_{\delta \sigma}$ but neither $G_{\delta}$ nor $F_{\sigma}$.)