$$(-1)^k\zeta^{(k)}(m)=\sum_{n=1}^\infty\frac{\ln^kn}{n^m}\approx\int_1^\infty\frac{\ln^kx}{x^m}dx=\frac{k!}{(m-1)^{k+1}}\qquad=>\qquad\zeta^{(k)}(0)\approx-k!$$

Indeed, the results can be checked numerically for $k\in\mathbb N^*$. So a meaningful value for $\zeta^{(1/2)}(0)$ could be something around $-\dfrac{\sqrt\pi}2$ . But, then again, $\zeta^{(0)}(0)=\zeta(0)=-\dfrac12$ is not a very good approximation of $-0!=-1$, so, for values of k in between $0$ and $1$, the approximation may lack relevance. Nevertheless, I am quite confident that a value somewhere in between $-\dfrac{\sqrt\pi}2$ and  $-\dfrac{\sqrt\pi}4$ is most likely. Hope this helps.

$(-1)^k\zeta^{(k)}(m)=\displaystyle\sum_{n=1}^\infty\frac{\ln^kn}{n^m}\approx\int_1^\infty\frac{\ln^kx}{x^m}dx=\frac{k!}{(m-1)^{k+1}}\quad=>\quad\zeta^{(k)}(0)\approx-k!$

Indeed, the results can be checked numerically for $k\in\mathbb N^*$. So a meaningful value for $\zeta^{(1/2)}(0)$ 

could be something around $-\dfrac{\sqrt\pi}2$ . But, then again, $\zeta^{(0)}(0)=\zeta(0)=-\dfrac12$ is not a very good 

approximation of $-0!=-1$, so, for values of k in between $0$ and $1$, the approximation may lack 

relevance. Nevertheless, I am quite confident that a value somewhere in between $-\dfrac{\sqrt\pi}2$ and 

$-\dfrac{\sqrt\pi}4$ is most likely. Hope this helps.