Note that $E[\varepsilon_n] = E[E[\varepsilon_n \mid X]] = 0$. Let $\sigma^2$ denote the variance of $\varepsilon_n$.

Let $Y_n = X + \varepsilon_1 + \dots + \varepsilon_n$. Note that $\frac{\varepsilon_1 + \dots + \varepsilon_n}{\sigma \sqrt{n}} \Rightarrow N(0,1)$ in distribution, by the central limit theorem, and $\frac{X}{\sigma\sqrt{n}} \to 0$ almost surely. So by Slutsky's theorem, $\frac{Y_n}{\sigma\sqrt{n}} \Rightarrow N(0, 1)$. Then $P(Y_n \le 0) = P(\frac{Y_n}{\sigma \sqrt{n}} \le 0) \to \Phi(0) = \frac{1}{2}$ where $\Phi$ is the standard normal cdf.

I think something similar should work if the $\varepsilon_n$ have infinite variance, using a different normalization and convergence to a stable law. I leave that as an exercise.

Note that $E[\varepsilon_n] = E[E[\varepsilon_n \mid X]] = 0$. Let $\sigma^2$ denote the variance of $\varepsilon_n$.

Let $Y_n = X + \varepsilon_1 + \dots + \varepsilon_n$. Note that $\frac{\varepsilon_1 + \dots + \varepsilon_n}{\sigma \sqrt{n}} \Rightarrow N(0,1)$ in distribution, by the central limit theorem, and $\frac{X}{\sigma\sqrt{n}} \to 0$ almost surely. So by Slutsky's theorem, $\frac{Y_n}{\sigma\sqrt{n}} \Rightarrow N(0, 1)$. Then $P(Y_n \le 0) = P(\frac{Y_n}{\sigma \sqrt{n}} \le 0) \to \Phi(0) = \frac{1}{2}$ where $\Phi$ is the standard normal cdf.