The sequence, as you have guessed right, converges to $2+3i$. Your argument is also almost fine. You need to clarify which norm you are using to prove convergence. Usually one uses the Euclidean norm, in which case, you get \begin{align} \left \lVert x_n - (2+3i) \right \rVert_2 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_2\\ & = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_2\\ & = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_2\\ & = \sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} \end{align} Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $\sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} < \sqrt{\left(\frac{3}{n+1} \right)^2 + \left(\frac{3}{n+1} \right)^2} = \frac{3 \sqrt{2}}{n+1}$ Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $\dfrac{3 \sqrt{2}}{n+1} < \epsilon.$ Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $\lVert x_n - \left( 2+3i\right)\rVert_2 < \epsilon$ Hence, $x_n \rightarrow 2+3i$.
EDIT
You could also use the $1$-norm as you have. \begin{align} \left \lVert x_n - (2+3i) \right \rVert_1 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_1\\ & = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_1\\ & = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_1\\ & = \left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert \end{align} Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $\left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert < \left \lvert \left( \frac{3}{n+1} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert = \frac{6}{n+1}$ Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $\dfrac{6}{n+1} < \epsilon.$ Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $\lVert x_n - \left( 2+3i\right)\rVert_1 < \epsilon$ Hence, $x_n \rightarrow 2+3i$.