Answer
Absolutely convergent.
Work Step by Step
Given $$
\sum_{n=1}^{\infty} \frac{(-1)^{n} e^{1 / n}}{n^{3}}
$$
Use the Limit Comparison Test with $a_n =\dfrac{ e^{1 / n}}{n^{3}}$ and $b_n=\dfrac{1}{ n^3}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^3 e^{1 / n}}{n^{3}}\\
&=\lim _{n \rightarrow \infty} e^{1 / n}\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{ n^3}$ is convergent ($p-$series with $p>1$), then $\displaystyle\sum_{n=1}^{\infty} \frac{ e^{1 / n}}{n^{3}}$ is also convergent, hence $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n} e^{1 / n}}{n^{3}}$ is absolutely convergent.