Thanks a lot @Derek for your argument which provides a logarithmic lower bound. I find it amazing that the order of magnitude of this (a priori) crude lower bound is the right one

Hmm, I did not think much about it, but I find it counter-intuitive for the least.

Thanks a lot Stefan for the example. Now I can see why I could not prove it. So what kind of lower bound can you get as a function of $n$. Is $\sqrt{n}$ good enough?

Dear Robert, I just had a look at Kuga's book. Basically, the answer to my question is positive if and only if the monodromy representation is triangulizable. Unfortunately, the book only proves the if direction (which is quite easy) and says that the only if direction is beyond the scope of the book. Do you know a reference off-hand where I could find a proof for the only if direction?

Dear Alexandre, I agree that my problem setup is vague and as you pointed out, ill-defined. So I just had a look at Kuga's book pages 109-115 (mentioned by Robert) and it seems that one may formulate precisely what I was intending to in my question. Any way, thanks a lot for pointing out the important distinction that one should make between a formal solution versus a convergent one; I had not thought about that.

Thanks a lot Igor, I did not know about this algorithm.

Dear Alexandre, see my added comment

Dear GH, yes you are right. Sorry for not having understood the point of your comment. So as you wrote, one may write down the appropriate second order linear differential equation and then find its general solution. Another way to see it is to notice that at $s=1/2$ there is a cancellation of the poles of $\zeta(2s)$ and $\zeta(2-2s)$ so that the next leading term turns out to be precisely $\log(y)\sqrt{y}$. Thanks for being persistant!