Fourier transform suggests that the limit function of $f_n(t)$ is constant $\pi$ (near $t=0$). But we cannot use the inversion theorem because $\sin(x)/x$ is not an $L^1$-function.
But it seems very plausible that $f_n(t)$ converges uniformly near $t=0$.To see this, draw some graphs of several $f_n$'s. But I don't know how to prove the uniform convergence. Any help?
First, exploiting the odd symmetry of $\frac{\cos((t-1)x)-\cos((t+1)x)}{x}$, we can write
$$\begin{align} \int_{-n}^n \frac{\sin(x)}{x}e^{-itx}\,dx&=\frac1{2}\int_{-n}^n \frac{\sin((t+1)x)-\sin((t-1)x)}{x}\,dx\\\\ &=\int_{0}^{n(t+1)}\frac{\sin(t)}{t}\,dt-\int_{0}^{n(t-1)}\frac{\sin(t)}{t}\,dt\tag1 \end{align}$$
Let $\delta>0$ and restrict $t$ such that $-1+\delta<t<1-\delta$. Then, we rewrite $(1)$ as
$$\begin{align} \int_{-n}^n \frac{\sin(x)}{x}e^{-itx}\,dx&=\int_{0}^{n(t+1)}\frac{\sin(t)}{t}\,dt+\int_{0}^{n(1-t)}\frac{\sin(t)}{t}\,dt\\\\ &=\pi-\int_{n(t+1)}^\infty\frac{\sin(t)}{t}\,dt-\int_{n(1-t)}^\infty\frac{\sin(t)}{t}\,dt \end{align}$$
Since the integral $\int_0^\infty \frac{\sin(x)}{x}\,dx$ converges, then for any $\varepsilon>0$ there exists a number $L>0$ such that $\ell>L$ implies
$$\left|\int_\ell^\infty \frac{\sin(x)}{x}\,dx\right|<\varepsilon/2$$
Therefore, whenever $n>L/\delta$ we have
$$\left|\int_{-n}^n \frac{\sin(x)}{x}e^{-itx}\,dx-\pi\right|<\varepsilon$$
which establishes the uniform convergence for $-1+\delta<t<1-\delta$ as was to be shown!