If $B_t$ is a Brownian motion, and a one-sided boundary stopping time is given by: $\tau_a=\inf\{t:B_t=a\}$ the distribution of $\tau_a$ is given by: $f_{\tau_a}(t)=\frac{|a|}{\sqrt{2\pi t^3}}\exp(-a^2/2t)$. In this case $\mathbb{P}(\tau_a>1)=1-\int_0^1f_{\tau_a}(t)dt$.
Now, for the two-sided boundary case. Define $\tau=\inf\{t:B_t=-\gamma \text{ or } B_t=\gamma\}$. Is there a method to calculate $\mathbb{P}(\tau>1)$? I only need that $\mathbb{P}(\tau>1)>0$.
In case that $\gamma \geq 1$, the claim follows from Wald's identities which state that
$$\begin{align*} \mathbb{E}\tau &= \gamma^2 \tag{1} \\ \mathbb{P}(B_{\tau}=\gamma) &= \mathbb{P}(B_{\tau}=-\gamma) = \frac{1}{2}. \tag{2} \end{align*}$$
If $\gamma \geq 1$, then $(1)$ clearly shows $\mathbb{P}(\tau>1)>0$. Using the scaling property, it is not difficult to see that $\mathbb{P}(\tau> \sqrt{\gamma}) > 0$ for any $\gamma>0$. Proving $$\mathbb{P}(\tau>1)>0$$ is much more delicate (... or, perhaps, I missed something obvious).
The idea of the proof is roughly the following: If the Brownian motion $(B_t)_{t \geq 0}$ satisfied $\mathbb{P}(\tau \leq 1)=1$, then it would grow too fast. The proof aims at constructing suitable stopping times and processes which enable us to find lower and upper bounds for the growth of $(B_t)_{t \geq 0}$.
Suppose that $\mathbb{P}(\tau \leq 1)=1$. We define iteratively Brownian motions in the following way: Starting with the given Brownian motion $(B_t)_{t \geq 0}$, we stop the Brownian motion when it leaves the interval $(-\gamma,\gamma)$ and start a new independent Brownian motion:
$$ \begin{align*} W_t^1 &:= B_t \\ W_t^2 &:= B_{\tau+t}-B_{\tau} \\ W_t^3 &:= W_{\tau^2+t}^2-W_{\tau^2}^2 \\ \vdots & \\ W_t^n &:= W_{\tau^{n-1}+t}^{n-1}- W_{\tau^{n-1}}^{n-1}. \end{align*}$$
where $\tau^n := \inf\{t \geq 0; |W_t^n| \geq \gamma\}$ and $$t_n := \sum_{j=0}^n \tau_j, \qquad n \geq 1.$$ (Note that $t_n = t_n(\omega)$ and $t_n \leq n$ almost surely.) In particular,
$$B_{t_n} = \sum_{j=1}^{n} W_{\tau_j}^j. $$ From $(2)$ we know that $$\mathbb{P}(W_{\tau^j}^j = \gamma) = \mathbb{P}(W_{\tau^j}^j = - \gamma) = \frac{1}{2}.$$
Since the Brownian motions $(W_t^j)$ are independent, we get
$$\begin{align*} \mathbb{P} \left( \sup_{s \leq n} B_s \geq (2k-n) \gamma \right) &\geq \mathbb{P}(B_{t_n} = (2k-n) \gamma) \\ &=\mathbb{P}(B_{t_n} = k \gamma + (n-k) (-\gamma)) \\ &= \binom{n}{k} \left( \frac{1}{2} \right)^k \left( \frac{1}{2}\right)^{n-k} =\binom{n}{k} \frac{1}{2^n}. \end{align*}$$
Now we replace $n \rightarrow r n$ and $k \rightarrow qr$ for $q > r/2$, $q,r \in \mathbb{N}$. Then the previous calculation shows
$$\mathbb{P}\left(\sup_{s \leq rn} B_s \geq n\gamma (2q-r) \right) \geq \binom{rn}{qn} \frac{1}{2^{qn}}.$$
Using Stirling's formula, one can show (see e.g. this paper or Wikpedia for the case $q=1$) that
$$\mathbb{P}\left(\sup_{s \leq rn} B_s \geq n\gamma (2q-r) \right) \geq \frac{c}{\sqrt{n}} \frac{r^{nr}}{q^{qn} (r-q)^{(r-q)n}} \frac{1}{2^{rn}}$$
for some constant $c>0$. This implies that for any $\delta<1$, we can choose $q,r$ (such that $\frac{q}{r}$ is close to $\frac{1}{2}$) such that
$$\mathbb{P}\left(\sup_{s \leq rn} |B_s| \geq n \gamma (2q-r) \right) \geq \frac{C}{\sqrt{n}} \delta^{rn} \tag{4}$$
for all $n \in \mathbb{N}$. On the other hand, using that $\sup_{s \leq t} B_s \sim |B_t|$, it is not difficult to see that
$$ \mathbb{P}\left( \sup_{s \leq t} B_s \geq x \right) \leq c \frac{\sqrt{t}}{x} \exp \left(- \frac{x^2}{2t} \right).$$
Consequently,
$$\begin{align*} \mathbb{P} \left( \sup_{s \leq rn} B_s \geq n \gamma (2q-r) \right) &\leq \frac{\tilde{c}}{\sqrt{n}} \exp \left(- \gamma^2 (2q-r)^2 \frac{r}{2} n \right) \\ &= \frac{\tilde{c}}{\sqrt{n}} \bigg[ \underbrace{\exp \left(-\gamma^2 \frac{(2q-r)^2}{2} \right)}_{\delta_0} \bigg]^{rn}. \end{align*}$$
Obviously, this contradicts $(4)$. (Choose $\delta > \delta_0$ and $n$ sufficiently large.)
Remark. Alternatively, the proof of the statement can be based on Lévy's triple law (see e.g. Brownian motion - An Introduction to Stochastic Processes by René Schilling/Lothar Partzsch for a proof).