It's a curious fact that for $n>0$, $\zeta^{(n)}(0)\approx -n!$. Apostol gave a table for $\frac{\zeta^{(n)}(0)}{n!}$, among other results on $\zeta^{(n)}(0)$ . the sequence :
$$\delta_{n}=\left | \zeta^{(n)}(0)+n! \right|$$
seems to be fast decreasing. What is the upper bound of $\delta_{n}$ ??
Edit: Following the logic in Apostol's paper, $\zeta(s)-\frac{1}{s-1}$ is holomorphic. Thus:
$$\zeta(s)-\frac{1}{s-1}=:A(s)=\sum_{n=0}^{\infty}\frac{A^{(n)}(0)}{n!}s^{n}$$ where : $$\left|A^{(n)}(0)\right|=\delta_{n}=\left|\zeta^{(n)}(0)+n!\right|$$ the expansion converges everywhere. Therefore : $$\limsup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n!}\right)^{\frac{1}{n}}=0$$ To be exact, I am interested in the limit : $$\limsup_{n\rightarrow\infty}\left(\frac{\delta_{n}}{n}\right)^{\frac{1}{n}}$$ hence the question !!



This should go as another comment to @Raymond Manzoni.
Here is a short routine in Pari/GP how the above coefficients can be computed to high accuracy by a very simple procedure:
The first 12 coefficients
$ \small \begin{matrix} 0.500000000000 \\ 0.0810614667953 \\ -0.00635645590858 \\ -0.00471116686225 \\ 0.00289681198629 \\ -0.000232907558455 \\ -0.000936825130051 \\ 0.000849823765002 \\ -0.000232431735512 \\ -0.000330589663612 \\ 0.000543234115780 \\ -0.000375493172907 \\ \vdots \end{matrix} $
and that around k=256 see Raymond's answer. Possibly we should increase the internal num-precision even higher to get meaningful digits below the decimal point for that high coefficients.
The computation to 120 good coefficients took only a few seconds with that given precision of 256 dec digits . For 256 good coefficients we need decimal precision \p 400 and much more memory and a couple of seconds more time
obsolete due to update dez 16 having the most simple precudere by "Vec()"
Pari/GP-script for "polcoeffs"