Berndt showed that the number of zeros of $\zeta^{(k)}(s)$ for $0<t<T$ is $$ N_k(T)=\frac{T}{2\pi}\left(\log\left(\frac{T}{4\pi}\right)-1\right)+O\left(\log T\right). $$ From this I think it follows that $\zeta^{(k)}(s)$ has the same genus and hence the same order as $\zeta(s)$.

Berndt showed that the number of zeros of $\zeta^{(k)}(s)$ for $0<t<T$ is $$ N_k(T)=\frac{T}{2\pi}\left(\log\left(\frac{T}{4\pi}\right)-1\right)+O\left(\log T\right). $$ From this I think it follows that $\zeta^{(k)}(s)$ has the same genus and hence the same order as $\zeta(s)$.