Yes, it is normal. To see it, write $f(re^{i \theta}) = \sum_{n=0}^{\infty}a_nr^ne^{in\theta}$ Calculate the integral by writing $|f|^2 = f \bar{f}$, and we obtain the inequality $\sum_{n=0}^{\infty}\frac{|a_n|^2}{n+1} \leq \frac{1}{\pi}$
Thus, if $f \in F$, $f(z) = \sum_{n=0}^{\infty}a_nz^n$, $|z| \leq r <1$ we have $|f(z)| \leq \sum_{n=0}^{\infty}|a_n|r^n = \sum_{n=0}^{\infty}\frac{|a_n|}{\sqrt{n+1}} \sqrt{n+1} r^n \leq (1/\pi)^{1/2} (\sum_{n=0}^{\infty} (n+1)r^{2n})^{1/2}$
and so $F$ is uniformly bounded on compact subsets, thus normal by Montel's theorem.