The problem is nontrivial already in the finite dimensional case $E= \mathbb R^d$, $F=\mathbb R$. The space $C^{\omega}(\Omega)$ of real-valued real analytic functions on the open bounded set $\Omega\subset \mathbb R^d$ does not have any obvious or natural metric which would make it a Fréchet space.
The good news is that there is a "canonical" topology which renders $C^{\omega}(\Omega)$ as a complete (reflexive nuclear separable) space. In fact, it is natural to endow $C^{\omega}(\Omega)$ with either an inductive limit or a projective limit topology but these two are equivalent on $C^{\omega}(\Omega)$ as was shown by Martineau in 1966.
For practical purposes, the topology can be described following the suggestion of Piero D'Ancona in his comment above. Let $\{U_j\}_{j\in\mathbb N}$ be a monotonically decreasing sequence of open sets of $\mathbb C^d$ such that $\Omega=\bigcap U_j$. Let $\{h_j\}_{j\in\mathbb N}$ be a sequence of bounded holomorphic functions $h_j:U_j\to\mathbb C$ such that $h_j|_{U_k}=h_k$ for $k\geq j$. Then a subbase element of the topology on $C^{\omega}(\Omega)$ has the form
$$\mathcal V_{j, K}=\left\{f\mbox{ is real analytic on }\Omega:\ \sup_{x\in K} \left|\partial^{\alpha} f\right|\leq C_j[\delta_j(K)]^{-|\alpha|}\ \mbox{ for every }\alpha\in\mathbb N^{d}_{0}\right\},$$
where the set $K\subset\Omega$ is compact, $\delta_j(K)=\mbox{dist}\{K,\partial U_{j+1}\}$ and $C_j$ is a constant which depends on the supremum of $h_j$ on $U_{j+1}$.
A sketch of the construction in the finite dimensional setting can be found, for instance, in A Primer of Real Analytic Functions by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way.
[EDIT. Concerning your specific question whether the limit of a sequence of real analytic functions is itself an analytic function. Let $f\in C^\infty(\mathbb T)$ be a periodic smooth but non-analytic function. Then the partial Fourier sums $S_N f$ converge to $f$ in the uniform metric with all their derivatives.]
Recently I found a counterexample to a somewhat stronger version of the question I originally asked, but which I actually needed. Namely for a smooth map $f\colon X\to Y$ which is transversal to $\Lambda\subset T^*Y\backslash 0$ (see e.g. Hormander's book for the definitions) the map $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}_{f^*\Lambda}(X)$ may not be topologically continuous (notice the subscript $f^*\Lambda$ in the target).
The counterexample is quite simple. Let $f\colon \mathbb{R}^2\to \mathbb{R}$ be the projection to the first coordinate. Let $\Lambda=\mathbb{R}\times (\mathbb{R}\backslash 0)$. Then $C^{-\infty}_{\Lambda}(\mathbb{R})=C^{-\infty}(\mathbb{R})$ is the usual Schwartz space with the usual weak topology. Furthermore $f^*\Lambda=\{(x,y;\xi,0)\}$. Then the map $f^*\colon C^{-\infty}_\Lambda(\mathbb{R})\to C^{-\infty}_{f^*\Lambda}(\mathbb{R}^2)$ is not topologically continuous.
Strictly speaking I do not yet have a counterexample to my original question, but this one already suffices my purposes.
Best Answer
More elementary than Ascoli:
If it was normable, it would mean that there exists a norm $n$ which is continuous, hence for some $k$, $n(\phi) \leqslant C\ p_k(\phi)$ and which defines the topology, i.e. such that for all $k$, $p_k(\phi) \leqslant C_k\ n(\phi)$.
This would imply that all the norms $p_k$ are equivalent for $k$ large enough, which is not the case: for example $\exp(-|x|^2) \sin(N x_1) /N^k$ tends to $0$ for the first $k$ norms but not for the following.