I tried the cases when $f(x)$ are the densities of normal and student t distribution. In both cases, the limit is $0$. I guess this conclusion might hold in general. I tried the following.
Let $a_n(x)=nf(x+n)$. We are interested in $\lim_{n\rightarrow\infty}a_n(x)$ if exists. For each $n$, we have $f(x+n)=a_n(x)/n$. We have $$ 1=\int_{-\infty}^\infty f(x+n)\,\mathrm{d}x=\int_{-\infty}^\infty \frac{a_n(x)}{n}\,\mathrm{d}x. $$ We can conclude at least that $a_n(x)<O(n)$, otherwise $\int_{-\infty}^\infty \frac{a_n(x)}{n}\,\mathrm{d}x$ is infinite.
My real target is $\lim_{n\rightarrow\infty}\int_{-\infty}^\infty nf(x+n)f(x)\,\mathrm{d}x$. Is this limit finite? Any thoughts about how to proceed would be appreciated.
Not always, and the integral itself need not be finite. For a counterexample, consider the PDF $$ f=\sum_{n\geqslant1}2^n\,\mathbf 1_{(n,n+1/4^n)}, $$ then, for every $t$, $$ \int_\mathbb Rf(x)f(x+t)\mathrm dx=\sum_{k,\ell}2^{k+\ell}\mathrm{Leb}((k,k+1/4^k)\cap(\ell-t,\ell+1/4^\ell-t)), $$ thus, for every nonnegative integer $n$, keeping only the indices $(k,\ell)$ such that $\ell=k+n$, one gets $$ \int_\mathbb Rf(x)f(x+n)\mathrm dx=\sum_{k=1}^\infty 2^{2k+n}\mathrm{Leb}((k,k+1/4^k))=+\infty. $$ For $C^\infty$ counterexamples, consider convolutions with normal densities of small variances.