You can't do it for
$n\geq 7$, since whichever die has the largest number wins at least $1/6$ of the time, which is too much.
$n=5$, since the state space has $6^5$ elements, which is not divisible by $5$.
It is possible for
$n=4$, using the dice $(24,17,13,10,7,4)$, $(23,18,14,12,5,1)$, $(22,19,15,11,8,2)$ and $(21,20,16,9,6,3)$.
I found these by greedily adding the largest numbers one by one to the die with the lowest probability so far, until some dice were sufficiently close to $1/4$ that I could determine what their other sides needed to look like.
Finally, it's not possible for
$n=6$. This is because all dice have to have a total of $7776$ winning combinations. Whichever die has number $36$ gets that many from that face alone, so if that die has exactly $7776$ winning combinations then none of the other faces on that die can possibly win. But then every face on every other die would have to be the winning score in a number of combinations that is a multiple of $5$ (possibly $0$ combinations, of course). This is because changing the outcome of the first die between the five small numbers makes no difference to which number wins. Thus if the first die has exactly $7776$ winning outcomes, none of the other dice can do so.
So the answer is
only $n=4$ is possible.