Any such $n$ gives a $(n+1)$-digit "permutable" or "absolute" prime. Clearly $n=0,1,2$ work, and it's conjectured that there are no absolute primes other than repunits larger than 991.
As hinted by @Sp3000, if 10 is a primitive root mod a prime $p$ then if $n>p$ then $p-1$ must divide $n+1$. According to this paper by Slinko, by considering all such primes up to $10^5$ Richert came up with a lower bound on the number of digits $>6\times 10^{175}$. The paper details which forms of numbers cannot be permutable primes, but notably multiple 1s and a single 3 is one of the few forms that remains as a possibility, implying that this case is still open.
Related, this article attributes the problem with several 1s and one 7 to Slinko and notes it as considered for the IMO. That version is resolved by Theorem 3 in the paper I cited above.