Show that the sequence of functions $\{f_n\}$ is not uniformly convergent on $\Bbb R.$

Here's an approach: note that $$ f_{n}(x) - f(x) = n \log \left (1 + \frac {x^2} {n} \right ) - \log(e^{x^2}) = \log \left( \frac{(1 + x^2/n)^n}{e^{x^2}} \right). $$ Now, we note that for any fixed $n$, we have $$ \begin{align} \lim_{x \to \infty} \frac{(1 + x^2/n)^n}{e^{x^2}} &= \left(\lim_{x \to \infty} \frac{1 + x^2/n}{e^{x^2/n}}\right)^n = 0 \end{align} $$ (since the numerator is a polynomial) from which we may conclude that (again, for any fixed $n$) the set $\{|f_n(x) - f(x)|: x \in \Bbb R\}$ is unbounded. Thus, if we fix (for instance) $\epsilon = 1$, we can see that for any $n$ there exists an $x$ such that $|f_n(x) - f(x)| \geq \epsilon$, which is to say that $\|f_n - f\|_{\sup} \geq \epsilon$.

So, the sequence is not uniformly convergent.

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Regarding your limit: with the substitution $m = \sqrt{1/n}$, we have $$ \begin{align} \lim_{m \to 0^+} 2m^{-3} \arctan \left (m \right ) &= 2\lim_{m \to 0^+}\frac{\arctan(m)}{m^3} = 2\lim_{m \to 0^+}\frac{\frac{1}{m^2 + 1}}{3m^2} = \frac 23 \lim_{m \to 0^+} \frac{1}{m^2(m^2 + 1)} = +\infty \end{align} $$

@mathmaniac.: (too big for a comment) using Taylor series, you rewrite $$n\log\Big(1+\frac{1}{n}\Big)-2n+2n\sqrt n\arctan\Big(\frac{1}{\sqrt{n}}\Big)$$ as $$n\Big(\frac{1}{n}+o\Big(\frac{1}{n}\Big)\Big)-2n+2n\sqrt{n}\Big(\frac{1}{\sqrt n}-\frac{1}{3(\sqrt n)^3}+o\Big(\frac{1}{(\sqrt n)^3}\Big)\Big)=1-\frac{2}{3}+o(1)\underset{n\to\infty}{\longrightarrow}\frac{1}{3},$$

so perhaps your mistake was to only use degree $1$ for the development of $\arctan$.