This is not a complete answer, but a continuation of the ideas in Christophe Leuridan's post. 

Write 

$$M_t := \underset{0 \leq s \leq t}{\text{max}} \, W_s.$$

Using the suggestion in the aforementioned post, we deduce the joint density $f_{(W_1, M_1)}$ of $(W_1, M_1)$ to be

$$f_{(W_1, M_1)} (x, y) = \sqrt{\frac{2}{\pi}} (2y - x) \,\text{exp} \left ( -\frac{1}{2}(2y - x)^2 \right ) \mathbf 1_{y > 0, x \leq y}.$$

We also know that $M_1$ has a half normal distribution, thus its probability density function is given by

$$f_{M_1} (y) = \sqrt{\frac{2}{\pi}} e^{-\frac{1}{2}y^2}.$$

Thus the conditional density $f_{W_1| M_1}$ of $W_1$ given $M_1$ is given by

$$f_{W_1| M_1} (x|y) = \frac{f_{(W_1, M_1)} (x, y)}{f_{M_1} (y)}$$

$$= (2y - x) \, \text{exp}\left (-\frac{1}{2}(3y^2 + x^2 - 4xy)\right) \, \mathbf 1_{y > 0, x \leq y}.$$

Hence we may compute

$$\lim_{\varepsilon \to 0+} \mathbb E_{\mathbb P^\varepsilon} (|X_1^\varepsilon - e|)$$

$$ = \lim_{\varepsilon \to 0+} \mathbb E_{\mathbb P^\varepsilon} (| e^{\varepsilon (W_1 - \frac{1}{2})} - e|)$$

$$= \lim_{\varepsilon \to 0+} \frac{\int_{\frac{1}{\varepsilon}}^\infty \int_{-\infty}^y |e^{\varepsilon (x - \frac{1}{2})} - e| \, f_{W_1| M_1} (x|y) \, dx \, dy}{\mathbb P(M_1 \geq \frac{1}{\varepsilon})}.$$

$$= \lim_{\varepsilon \to 0+} \frac{\int_{\frac{1}{\varepsilon}}^\infty \int_{-\infty}^y |e^{\varepsilon (x - \frac{1}{2})} - e| \, (2y - x) \, \text{exp}\left (-\frac{1}{2}(3y^2 + x^2 - 4xy)\right ) \, dx \, dy}{2 \Phi(\frac{1}{\varepsilon})}.$$

Where $1 - \Phi$ is the CDF of the standard normal distribution.

As of now, it is unclear to me how to evaluate the above limit.