Firstly, $m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T$. So, you can either consider the running maximum or minimum.
Let $\tau$ be a stopping time and $(B_t)$ a Brownian motion. Then,
\begin{align*}
W_t =\begin{cases}
B_t & t\leq \tau, \\
2B_\tau - B_t & t\geq \tau,
\end{cases}
\end{align*}
is again a standard Brownian motion (This is the reflection principle).
For $a\geq 0$ and $t>0$, the reflection principle implies that
\begin{align*}
\mathbb{P}[\{M_T\geq a\}] &= 2\mathbb{P}[\{B_t\geq a\}] \\
\implies \mathbb{P}[\{M_T\leq a\}] &= 2\mathbb{P}[\{B_t\leq a\}]-1 \\
&= 2\Phi\left(\frac{a}{\sqrt{t}}\right)-1.
\end{align*}
Thus, the probability density function of $(M_t)$ is given by
\begin{align*}
f_{M_t}(x) &= \frac{\partial }{\partial x} \left(2\Phi\left(\frac{x}{\sqrt{t}}\right)-1\right) \\
&= \frac{2}{\sqrt{t}}\varphi\left(\frac{x}{\sqrt{t}}\right) \\
&= \sqrt{\frac{2}{t\pi}}e^{-\frac{1}{2t}x^2}
\end{align*}
for $x\geq0$ and $f_{M_t}(x)=0$ for $x<0$. The function is clearly non-negative and you can easily see that it integrates to one.
Here, $\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$ is the pdf of a standard normally distributed random variable and $\Phi$ the corresponding cdf.