It is $\ell$-divisible for every odd number $\ell$.
To see this, let $\alpha \in Br(K)$, and look at its image in $(\alpha_\nu)_\nu \in\oplus_\nu \mathbb{Q}/ \mathbb{Z}$. You know that each component is divisible by $\ell$, so you can form, in several ways, the element $(\alpha_\nu /\ell)_\nu$. The problem is that now it might be that $inv((\alpha_\nu /\ell)_\nu)$ is not $0$.
denote this number by $d \in \mathbb{Q} / \mathbb{Z}$. This is actually an element of $(1/\ell) \mathbb{Z} / \mathbb{Z}$. We can choose some $\nu_0$ and modify
$(\alpha_\nu / \ell)_\nu$ by subtracting $d|_{\nu_0}$ (i.e. the element that has $d$ in the $\nu_0$ summand and 0 elsewhere) and get a new element
$(\beta_\nu)_\nu$. The difference between $(\alpha_\nu / \ell)_\nu$ and
$(\beta_\nu)_\nu$ is $\ell$-torsion, so clearly we still have
$\ell (\beta_\nu)_\nu = (\alpha_\nu)_\nu$, but on the other hand
$inv(\beta_\nu) = 0$ so it comes from an element $\beta$ of $Br(K)$.
Since the map $Br(K) \to \oplus_\nu Br(K_\nu)$ is injective, this implies
$\ell \beta = \alpha$.
Regarding the second question, $Br(K)[n]$ is not finite. For example, choose $\nu_0$ and consider all the elements of the form $(1/n)|_{\nu_0} - (1/n)|_{\nu}$ for $\nu$ a non-Archimedean place. Then it is an infinite sequence of different elements in it.