Here is what the solution in the textbook says:
Let $S_n=X_1+\cdots+X_n$, with $X_1,X_2,\ldots$ i.i.d. $\sim \mathrm{Pois}(1)$. Then $S_n \sim\mathrm{Pois}(n)$ and for $n$ large, $S_n$ is approximately $N(n,n)$ by the CLT.
Based on my understanding of the Central Limit Theorem, $\overline{S}_n$ is approximately $N(n,n)$, not $S_n$. Here is my proof:
The Central Limit Theorem says \begin{align} \sqrt{n}\left(\frac{\overline{S}_n-\mu}{\sigma}\right)\xrightarrow{n \rightarrow \infty}\mathcal{N}(0,1) \\ \sqrt{n}\left(\frac{\overline{S}_n-n}{n}\right)\xrightarrow{n \rightarrow \infty}\mathcal{N}(0,1) \\ \end{align}
With a location scale transformation we get:
\begin{align} \overline{S}_n-n \xrightarrow{n \rightarrow \infty}\mathcal{N}(0,n) \\ \overline{S}_n \xrightarrow{n \rightarrow \infty}\mathcal{N}(n,n) \\ \end{align}
And since $\overline{S}_n=\frac{1}{n}S_n$, $S_n \xrightarrow{n \rightarrow \infty}\mathcal{N}(n,n^3)$.
Is my proof incorrect or is the problem just a difference in notation?
The mean and standard deviation of a single summand are $1$, not $n$, so you have just $\sqrt{n}\left(\overline{S}_n-1\right)$ going to $N(0,1)$, which is consistent with the desired statement.
That said, the CLT does not really say things like "$S_n$ is approximately $N(n,n)$". It only really describes about shifted+rescaled distributions converging to fixed normal distributions.