Fourier transformation of the Toeplitz matrix elements in the infinite-matrix limit gives,
$$w(\theta)=\sum_{m=-\infty}^\infty \frac{e^{im\theta}}{1+im\epsilon}=\frac{i}{\epsilon} \left[e^{-i \theta} \Phi \left(e^{-i \theta},1,\frac{\epsilon+i}{\epsilon}\right)-\Phi \left(e^{i \theta},1,-\frac{i}{\epsilon}\right)\right],$$
with $\Phi$ the Hurwitz-Lerch transcendent. The Fourier transformed matrix is diagonal, so the estimate for the eigenvalues of $G$ for $n\gg 1$ is $$\lambda_k=w(2\pi k/n),\;\; k=1,2,\ldots n.$$
This is a monotonically decreasing fnction of $k$. The smallest eigenvalue is reached for $k=n$,
$$\lambda_{\rm min}=w(2\pi)=\frac{i}{\epsilon} \left[ \Phi \left(1,1,\frac{\epsilon+i}{\epsilon}\right)-\Phi \left(1,1,-\frac{i}{\epsilon}\right)\right].$$

A numerical check, for $n=100$, shows this is very accurate:

<IMG SRC="https://i.sstatic.net/RL5G1.png" WIDTH="400"/>

Red curve is $w(2\pi)$ as a function of $\epsilon$, the blue dots are the smallest eigenvalue of the $100\times 100$ Toeplitz matrix.