Let $T:=\inf\{t>0: |B_t|=1\}$ be a hitting time for standard Brownian motion. I want to show that $$\lim_{t\to\infty} e^{\frac{\pi^2}{8}t}\mathbb{P}[T\geq t]=\frac{4}{\pi}$$
I had a look at A bound for the probability that a Brownian motion stays in an interval and only managed to get a the bound $$\mathbb{P}[T\geq t]\leq e^\frac{-tu^2}{2}\frac{1}{\cos(u)}$$ for all $u\in[0,\frac{\pi}{2})$, but it's not quite enough.
Observe that $T=\min\{T_1, T_{-1}\}$ where $T_1$, respectively $T_{-1}$ denote the first hitting time of a Brownian motion $B$ at point $1$, respectively $-1$. Then
$$P(T>t)=P(\min \{T_{1}, T_{-1}\}>t) = P(T_1 >t, T_{-1}>t) = P(T_{-1} >t | T_{1}>t) P(T_1>t).$$
Then clearly $$P(T>t) \leq P(T_1>t).$$ Now we know that $T_1$ is Lévy distributed with $\mu = 0$ and $c=1^2$, see http://en.wikipedia.org/wiki/L%C3%A9vy_distribution.
Hence $$P(T>t) \leq P(T_1>t) = \frac{1}{\sqrt{2\pi}}\int_t^{\infty} s^{-3/2} e^{-\frac{1}{2s}}ds.$$
Now take $e^{\alpha t^{\beta}}$, $\alpha,\beta\in \mathbb{R}$. Hence $$e^{\alpha t^{\beta}}P(T>t) \leq \frac{1}{\sqrt{2\pi}} e^{\alpha t^{\beta}}\int_t^{\infty} s^{-3/2} e^{-\frac{1}{2s}}ds.$$
Then we are interested in finding $\alpha,\beta$ so that the following limit is a finite constant different from 0. So
$$\lim_{t\to \infty} \frac{ \int_t^{\infty} s^{-3/2} e^{-\frac{1}{2s}}ds }{e^{-\alpha t^{\beta}}} = \, "\frac{0}{0}"$$
Then using L'Hôpital's rule and fundamental theorem of calculus we have $$\lim_{t\to \infty} \frac{ \int_t^{\infty} s^{-3/2} e^{-\frac{1}{2s}}ds }{e^{-\alpha t^{\beta}}} =\frac{1}{\alpha \beta}\lim_{t\to \infty} \frac{t^{-3/2} e^{-\frac{1}{2t}} }{ t^{\beta-1}e^{-\alpha t^{\beta}}}.$$
Now if $\beta = -1/2$ then $$\lim_{t\to \infty} \frac{ \int_t^{\infty} s^{-3/2} e^{-\frac{1}{2s}}ds }{e^{-\alpha t^{\beta}}} = \frac{-2}{\alpha}.$$
In a summary we have
$$\lim_{t\to \infty} e^{\alpha t^{-1/2}}P(T>t) \leq \frac{1}{\sqrt{2\pi}} \frac{-2}{\alpha}$$ and you are free to choose the $\alpha$ you like.
I don't know if this estimate is too crude for your purposes. If you want exact limit I guess you have to compute $P(T_1>t |T_{-1}>t)$ explicitly and proceed in the same way.