For $n = 8$ we have $3n^2 + n + 1 = 201$ whose sum of the digits is $3$.

We prove that the sum of the digits of $3n^2+n+1$ cannot be $1$ or $2$. As $3n^2+n+1$ is odd, it cannot be written as $10^k$ or $2 \cdot 10^k$, $k \in \mathbb{N}$, nor can it be written as $10^k+10^j$ with $k, j > 0$. If $3n^2+n+1 = 10^k+1$, then $n(3n+1) = 10^k$. But $n$ and $3n+1$ are co-prime, hence they must be $2^k$ and $5^k$. As $n < 3n+1$, we must have $n = 2^k$ and $3n+1 = 5^k$, which is not possible because $5^k > 4^k > 3 \cdot 2^k + 1$ if $k \ge 2$, and $k=1$ does not work either.