The condition $r>\sqrt 2$ is also sufficient. The random variables $(M_n)_{n\geq1}$ are iid (in fact, we need that they are identically distributed), due to the independence of the increments and the translation invariance of Brownian motion. We shall use the first Borel-Cantelli lemma to prove that for $r>\sqrt 2$,
$$
\mathbb{P}[M_n\geq r\sqrt{\log n}\text{ infinitely often}]=0.
$$
To this end, it is sufficient to prove that
$$
\sum_n \mathbb{P}[M_n\geq r\sqrt{\log n}]<\infty.
$$
Indeed,
$$
\mathbb{P}[M_n\geq r\sqrt{\log n}]=\mathbb{P}[M_1\geq r\sqrt{\log n}]\leq 2\mathbb{P}[U_1\geq r\sqrt{\log n}],
$$
where $U_t=\max\{B_s:s\in[0,t]\}$ is the running maximum (the inequality above holds since $M_1$ is the largest between $U_1$ and $-L_1$, and $U_t$ and $-L_t$ have the same distribution, where $L_t$ is the running minimum). By the reflection principle,
$$
\mathbb{P}[U_1\geq r\sqrt{\log n}]=2\mathbb{P}[B_1\geq r\sqrt{\log n}]
=2(1-\Phi(r\sqrt{\log n})),
$$
and the series converges for $r>\sqrt 2$.