By the reflection principle, we can extend $f$ to the larger rectangle $R=[-1,1]\times [-2,2]$, and the upper bound $|f|\le 1$ holds for the extended function as well. The rectangle $R$ contains a disk of radius $1$ centered at $i/4$. Applying the Cauchy integral formula on this disk, we find that $|f^{(n)}(i/4)|\le n!$ for all $n$. In particular, $|f^{(5)}(i/4)|\le 120$, which I believe was the intent of the problem. For the 8th derivative we get $|f^{(8)}(i/4)|\le 40320$.
Indeed, I think I have a counterexample to the stated bound on $f^{(8)}(i/4)$. The idea is to map $R$ into the unit disk $\mathbb D$. Mapping rectangles is hard (elliptic functions, whatever). Instead I'm going to map a larger domain: the vertical strip $S=\{z:-1<\operatorname{Re}z<1\}$. This is a standard exercise with conformal maps: $\psi(z)= i\,\frac{\exp(\pi i z/2)-1}{\exp(\pi i z/2)+1}$ maps $S$ onto $\mathbb D$ and is real on the segment $[-1,1]$ (this segment is sent onto a half of the unit circle by the exponential map, and then onto horizontal diameter by the Möbius map).
No, $\psi$ is not my example. After all, $|\psi^{(8)}(i/4)|<24$ as you can find from your friendly computer algebra system, e.g., evalf(eval(diff(psi,z8),z=Pi*I/4))
in Maple. When maximizing the 8th derivative, we should take the 8-fold cover: f(z)=\psi(z)^8$. The same Maple thing now tells me that $|f^{(8)}(i/4)|>1800$.