jmc

Since $t(n)=11$ is prime and is the product of 1 more than each of the exponents in the prime factorization of $n$, there can be only one exponent, and therefore one prime in the prime factorization of $n$. This means $n=p^{10}$ for some odd prime number $p$, so $$8n^3=2^3\cdot p^{30}\Rightarrow t(8n^3)=(3+1)(30+1)=\boxed{124}.$$