You are on the right track: apply Lebesgue's dominated convergence with $g(x)=\frac{1}{1+x^2}$ which is Lebesgue integrable in $[0,+\infty)$. Since $f_n(x)=\frac{\sin(e^x) }{1+nx^2}\to 0$ for all $x>0$ the sequence $(f_n)_n$ converges to zero almost everywhere on $[0,+\infty)$, that's enough for dominated convergence, and we may conclude that the limit of $\int_0^{\infty} f_n(x)\,dx$ is zero.
Alternative way (without dominated convergence):
$$\begin{align}\left|\int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\, dx \right|&\leq \int_0^{+\infty} \frac{|\sin(e^x) |}{1+nx^2}\, dx\\
&\leq \int_0^{+\infty} \frac{dx}{1+nx^2}=\left[\frac{\arctan(\sqrt{n}x)}{\sqrt{n}}\right]_0^{+\infty}=\frac{\pi}{2\sqrt{n}}.\end{align}$$
So, again, the limit as $n\to\infty$ is zero.
Let $f(x) = \int_{x}^{\infty} \frac{\mathrm{d}t}{1+t^3}$. Then $f$ is integrable on $[0, \infty)$. Indeed, $f$ is bounded by $f(0)$ and $f(x) \asymp 1/x^2$ as $x\to\infty$. Now by integration by parts,
\begin{align*}
\int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x
&= -\int_{0}^{\infty} n^2(\cos(x/n^2)-1)f'(x) \,\mathrm{d}x \\
&= \underbrace{\left[ -n^2(\cos(x/n^2)-1)f(x) \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \sin(x/n^2)f(x) \, \mathrm{d}x.
\end{align*}
Now by the dominated convergence theorem,
$$ \lim_{n\to\infty} \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x
= - \int_{0}^{\infty} \lim_{n\to\infty} \sin(x/n^2)f(x) \, \mathrm{d}x
= 0. $$
Addendum. A more detailed analysis, with a bit of help from Mathematica 11, shows that
$$ \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x = -\frac{1}{n^2}\left(\log n + \frac{3}{4} - \frac{\gamma}{2} + o(1) \right) $$
as $n\to\infty$, where $\gamma$ is the Euler-Mascheroni constant.
Best Answer
Substitute $x=\frac{t}{\sqrt{n}}$
Then according to the control convergence theorem (it is not difficult to verify the applicable conditions),$$\operatorname*{lim}_{n\to\infty}\int_{0}^{+\infty}\frac{n\sin\frac{t^{2}}{n}}{(1+\frac{t^{2}}{n})^{n}}\mathrm{d}t=\int_{0}^{+\infty}\lim_{n\to\infty}\frac{n\sin\frac{t^{2}}{n}}{(1+\frac{t^{2}}{n})^{n}}\mathrm{d}t=\int_{0}^{+\infty}t^{2}e^{-t^{2}}\mathrm{d}t=\frac{\sqrt{\pi}}{4}.$$