Against the drift

Idea. Find an exponential of the process that is a martingale. Its value at the hitting times is known, and optional stopping turns that into an equation for the probability.

Step 1: the martingale. $\mathbb{E}[e^{\lambda X_t}]=e^{\lambda t+{\lambda }^{2}t/2}$. This equals $1$ for all $t$ when $\lambda +\frac{{\lambda }^2}{2}=0$, that is $\lambda =-2$. With independent increments, $M_t=e^{-2X_t}$ is a martingale with $M_0=1$.

Step 2: optional stopping. Let $\tau$ be the first time $X$ hits $-1$ or $b>0$. The drift makes $\tau$ finite, and $M$ is bounded up to $\tau$ (between $e^{-2b}$ and $e^2$). So $$1=\mathbb{E}[M_{\tau }]=p_b{e}^2+(1-p_b)e^{-2b},$$ where $p_b$ is the probability of hitting $-1$ before $b$.

Step 3: let $b\to \infty$. $p_b$ increases to the probability of ever hitting $-1$, and $e^{-2b}\to 0$, so $$\mathbb{P}(X\text{ ever reaches }-1)=e^{-2}\approx \mathrm{0.135.}$$

In general. For $X_t=\mu t+\sigma W_t$ with $\mu >0$, the probability of ever going down by $a$ is $e^{-2\mu a/{\sigma }^2}$. It is the continuous version of gambler's ruin against an infinitely rich opponent.