You might be able to construct lower bounds of losing probabilities.
If you have 1 sided dice, and the first dice has a value on that one side, if the second wins against the first it has to be higher than the first, and if the thrid wants to win against the second it has to be higher than the second. But as numbers are transitive the thrid will also always win against the first.
So there is at least one dice that has probability 1 to lose.
If you have 2 sided dice, you can chose only one side of those three dice for now. Again there is at least one dice that loses against both of the other dice if this side shows. Now you would give the other side of the losing dice a winning side against both dices to balance it out. But you can't lower the probability of losing below 1/2 (probability of throwing the first side on your dice) * 1/2 (probability of your opponent throwing the first side) so in total $1/2^2$.
Continuing on, you can already tell that there will always exist at least one dice of three n-sided dice that loses with at least $1/n^2$ probability. If you assume that your solution will results in 3 dice such that you will have the same probability of winning whatever dice your opponent picks this results in the fact that you will lose with at least $1/n^2$ probability with either dice.
And this was a very bad lower estimate. But you get the idea. If you don't just throw away all the interactions from the following sides you might get a better estimate. Look for repeating patterns and then maybe try induction.
Personally I would try to prove the hypothesis that the sum of losing probabilites over all dice must be greater than 1
Intuition: Given that you have the probability of 1 on one dice in the one sided case you already get that.
On the two sided dice you only get 1/4 probability on the second dice by assuming the first side will lose against both other first sides but always win against the second side - which would imply that the second side consists only of smaller numbers (because this dice has the smallest number of the first side). So if you then take the second side there is also a dice that is the lowest on the second side. So this dice loses against the two other dice if they throw the first side (because the first side is always higher than the second side - assumption). And it always loses when they throw the second side. So it has the probability of losing of 1. $1+1/4\ge1$. But if you construct the second side in such a way that it is not completely lower than the first side you will increase the propability of 1/4 to something higher. And again there must be a mirror dice for the second dice so I think you will get something higher than 1 always. But that is just my hypothesis.
If that hypothesis would be true, then 1/3 would be a lower bound to the losing probability of the dice if you distribute it evenly. You could try to prove that. Then you would also know that you can't get over a winning probability of 2/3.