Analytic formula and an exponential bound
For $r \in [0, 1]$, let $ \tau_m(r) := (1/\lambda(B_m(1)))^2 \int_{B_m(r)} \lambda(B_m(1) \cap (B_m(1) + x))dx$. It has been proven by other users that
$$ \tau_m(r) = \frac{1}{V_m^{cap}(0)}\int_{0}^r ms^{m-1}V_m^{cap}(s/2)ds, $$
where $V_m(h)$ is the volume (i.e Lebesgue measure) of the half-lens $\{x \in B_m \mid x_1 \ge h\}$. I propose to do the actual calculations and get an analytic formula in terms of special functions (beta, gamma, etc.).
The end result will be the following result on the exponential decay of $\tau_m(r)$ as a function on the dimension $m$.
Theorem. For every $r \in [0, 1]$ and large integer $m$, we have the bound $$ \tau_m(r) \le r^me^{-\frac{r^2}{8}} + e^{-m\frac{(1-r^2/2)^2}{2}}. $$ In particuar, when $r=1$ it holds that $$ \tau_m(1) \le 2e^{-m\frac{r^2}{8}}. $$
Proof.
Now, it is a classical computaiton that $V_m^{cap}(h) = \beta_{1-h^2}\left(\frac{m+1}{2},\frac{1}{2}\right)$, where $\beta_x(a, b) := \int_0^xt^{a-1}(1-t)^{b-1}dt$ defines the incomplete beta function. In particular, $V_m^{cap}(0) = \beta_{1}\left(\frac{m+1}{2},\frac{1}{2}\right)$.
Thus, integrating by parts, we get
$$ \begin{split} V_m^{cap}(0)\tau_m(1) &= \int_{0}^r s^{m-1}V_m^{cap}(s/2)ds = \int_{0}^r \beta_{1-s^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)ds^m\\ &= \left[s^m\beta_{1-s^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)\right]_0^r + \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds \\ &= r^m \beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right) + I_m, \end{split} \tag{1} $$ where $I_m := \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds$. Now, consider the change of variable $s = 2\sqrt{t}$. This gives $s^m(1-s^2/4)^{(m-1)/2}ds=2^mt^{n/2}\cdot(1-t)^{(m-1)/2}\cdot s^{-1/2}ds = 2^mt^{(m-1)/2}(1-t)^{(m-1)/2}$. Thus,
$$ \begin{split} I_m &= \int_0^r s^m(1-s^2/4)^{\frac{m-1}{2}}ds = 2^m\int_0^{r^2/4}t^{(m-1)/2}(1-r)^{(m-1)/2}dt\\ &=: 2^m\beta_{1/4}\left(\frac{m+1}{2},\frac{1}{2}\right). \end{split} \tag{2} $$
Combining (1) and (2) then gives the analytic formula,
Analytic formula. $$ \tau_m(r) = \frac{r^m\beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right) + 2^m\beta_{r^2/4}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{m\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} $$
Now, by well-known sub-Gaussian concentration inequalities for the beta distribution, one has
$$ \begin{split} \frac{\beta_{1-r^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} &= \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{1}{2}} \le 1-r^2/4\right) = \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{1}{2}} - 1 \le -r^2/4\right)\\ &\le e^{-(r^2/4)^2\cdot \frac{1}{2\cdot 1/(2m)}} = e^{-mr^4/16}. \end{split} $$
Likewise,
$$ \begin{split} \frac{\beta_{r^2/4}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_{1}\left(\frac{m+1}{2},\frac{m+1}{2}\right)} &= \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{m+1}{2}} \le r^2/4\right) = \mathbb P\left(\mathrm B_{\frac{m+1}{2},\frac{m+1}{2}} - \frac{1}{2} \le -\frac{1-r^2/2}{2}\right)\\ &\le e^{-((1-r^2/2)/2)^2/2\cdot 4(m+2)} = e^{-(m+2)(1-r^2/2)^2/2} \le e^{-\frac{m(1-r^2/2)^2}{2}}. \end{split} $$
On the other hand, setting $\alpha_m = (m+1)/2$ for large $m$, Stirling's formula gives
$$ \beta_{1}\left(\alpha_m,\alpha_m\right) \approx \sqrt{2\pi}\frac{\alpha_m^{2\alpha_m-1}}{(2\alpha_m)^{2\alpha_m-1/2}} = \frac{\sqrt{\pi}}{2^{2\alpha_m-1}} = \frac{1}{2^m} \sqrt{\frac{2\pi}{m+1}}\approx \frac{1}{2^m}\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right) $$
Putting things together gives
$$ \tau_m(r) \le r^me^{-m\frac{r^2}{8}}+e^{-m\frac{(1-r^2/2)^2}{2}}. $$