When $n=1$, we have $a_1=S_1=1+1+1=3$.

For $n\geqslant 2$, we have $a_n=S_n-S_{n-1}=n^2+n+1-[(n-1)^2+(n-1)+1]=2n$.

Therefore, when $n=1$, it does not satisfy the above formula.

Thus, the general formula for the sequence $\{a_n\}$ is $a_n= \begin{cases} 3, & n=1 \\ 2n, & n\geqslant 2 \end{cases}$.

Hence, the answer is: $a_n= \begin{cases} 3, & n=1 \\ 2n, & n\geqslant 2 \end{cases}$.

By using "When $n=1$, $a_1=S_1$; for $n\geqslant 2$, $a_n=S_n-S_{n-1}$", we can derive the solution.

This problem examines the relationship between the sum formula of the first $n$ terms of a sequence and its general term formula. Mastering "When $n=1$, $a_1=S_1$; for $n\geqslant 2$, $a_n=S_n-S_{n-1}$" is key to solving the problem.

Therefore, the final answer is $\boxed{a_n= \begin{cases} 3, & n=1 \\ 2n, & n\geqslant 2 \end{cases}}$.