Here is a hint on how to do the problem. Because $\phi_n(x) = 0$ for $|x| \geq 1/n$, you can say
$\int_{-1}^1 \phi_n(x) dx = \int_{-\infty}^{\infty} \phi_n(x) dx.$
Now choose $x \in\Bbb{R}$. Then
$\begin{eqnarray*} |f_n(x) - f(x) | &=& \left|\lim_{N \to \infty} \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy \right| \\ &=&\lim_{N \to \infty} \left| \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy \right| \\ &\leq&\lim_{N \to \infty} \int_{-N}^N |\phi_n(x-y)||f(y) - f(x)| dy \\ &=& \lim_{N \to \infty} \int_{x-N}^{x+N} |\phi_n(u)||f(x) - f(x-u)| du \\ &=& \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)||f(x) - f(x-u)| du. \end{eqnarray*}$
Facts I used above: The absolute value function is continuous, and the integrals above are continuous functions in $x$. Also, I used that the $\phi_n(x)$ are zero for $n$ sufficiently large, and there is a change of variables above. We can now complete your problem as follows. By continuity of $f$ we know that given any $\epsilon > 0$ , there is $\delta > 0$ such that $|u| < \delta$ will imply that $|f(x) - f(x-u)| < \epsilon$. Choose $N$ so large that $1/N < \delta$. Then for all $n\geq N$, we would have that $|u| < \frac{1}{n}$ will imply that $|f(x) - f(x-u)| < \epsilon$. The integral in the last line is now less than
$ \epsilon \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)| du = C\epsilon$
for some constant $C$. It follows that $f_n \to f$ pointwise.