Let $X_i(t)$ be standard Brownian motion in 1D. Define the radial Brownian motion as $\displaystyle R(t) = \sqrt{X_1(t)^2 + \cdots + X_N(t)^2}$.
How do we lower bound the probability $\mathbb{P}(R(T)<r \,\, \textrm{for all }0<t<T)$ for a fixed $T$ and $r$?
I do not know if you need a more refined answer. This one is a simple estimate I would be curious to see more advanced approaches.
An easy lower bound could be this. I call $M_X(t)$ the maximum between 0 and t of the modulus of the random variable $X$. Than:
$\mathbf P(M^2_R(t)\le r^2)>\mathbf{P}(M_{X_i(t)}^2\le \frac{r^2}{N}, i=1,...N)=\prod_{i=1}^N \mathbf{P}(M_{X_i(t)}^2\le \frac{r^2}{N})=(\mathbf{P}(M_{B(t)}^2\le \frac{r^2}{N}))^N$
where $B$ is a standard Brownian motion. The idea is that if each of the Brownian motions individually is smaller than $r^2/N$ than their sum is smaller than $r^2$. Of course this is just a sufficient condition therefore it is only a lower bound. The analytical form of the distribution of the maximum of B can be found in books.