Given $f\in C^{\infty} (E)$, where $E\subseteq \mathbb{C}$, define $E_{\rho} \subseteq \mathbb{C}$ as the maximal ellipse with foci at $\{-1,1\}$ where $f$ is analytic, and semi-minor + semi-major axis summing to $\rho$.
Question: Is there any connection between $\rho$ and the modulus of continuity $\omega (\delta )$ of $f$ in $\lbrack -1, 1 \rbrack$?
There is indeed a relation between $\rho$ and the modulus of continuity $\omega_{f}$ of $f$ on $[-1,1]$ which is obtained via the rate of polynomial approximation to $f$ on $[-1,1]$. Denote by $E_{n}(f)$ the distance from $f$ to polynomials of degree at most $n$ with respect to the uniform norm on $[-1,1]$. By the classical Jackson's theorem, $$E_{n}(f)\leq C\omega_{f}(1/n),$$ with $C$ some constant. On the other hand, by the Bernstein-Walsh theorem (see e.g. the book Potential theory in the complex plane by T. Ransford for a reference), one has $$\limsup_{n}E_{n}(f)^{1/n}= e^{-\inf_{\mathbb{C}\setminus E_{\rho}}g(z,\infty)},$$ where $g(z,\infty)$ denotes the Green function of $\mathbb{C}\setminus[-1,1]$ with pole at infinity. Since the minimum of $g(z,\infty)$ is attained on the boundary of the ellipse $E_{\rho}$ and equals the constant value $\log\rho$ there, one derives that $$\frac1\rho\leq\limsup_{n}\omega_{f}(1/n)^{1/n}.$$ In other words, a function with an "asymptotically" small modulus of continuity should admit an analytic continuation to a "large" ellipse $E_{\rho}$. Regarding your example (in the comment) of a narrowing gaussian, the left-hand side in the above inequality vanishes so nothing can be derived about its modulus of continuity.
