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. (Beware of a $2\pi/\epsilon$ discontinuity at $\theta=0$ mod $2\pi$.)
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 function of $k$. For the smallest eigenvalue I find $$\lim_{n\rightarrow\infty}\lambda_{\rm min}=\lim_{\theta\uparrow 2\pi}w(\theta)=\frac{\pi}{\epsilon} \left(\coth \left(\frac{\pi }{\epsilon}\right)-1\right).$$
A numerical check, for $n=100$, shows this is quite accurate:
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.
Numerically, I find that for finite $n$ the smallest eigenvalue remains above the large-$n$ limit, indicating that $(\pi/\epsilon)[\coth(\pi/\epsilon)-1]$ is a lower bound (but I have no proof for that).
Above I compared the smallest eigenvalue, the full set of eigenvalues also agrees nicely with $w(2\pi k/n)$. Here is a comparison for $\epsilon=1$ and $n=100$.
Red curve: $w(\theta)$ for $\epsilon=1$; blue data points: $\lambda_k$ as a function of $2\pi k/n$ for $n=100$.
For the $\epsilon$-dependence of the largest eigenvalue $\lambda_{\rm max}$ I find $$\lim_{n\rightarrow\infty} \lambda_{\rm max}=\lim_{\theta\downarrow 0}w(\theta)=\frac{\pi}{\epsilon} \left(\coth \left(\frac{\pi }{\epsilon}\right)+1\right).$$ Here is a comparison of $w(0)$ with $\lambda_{\rm max}$ for $n=500$:
