Given a random variable $X ~ Norm[μ,σ]$ (a normal distribution with mean $\mu$ and standard deviation $σ$ and a constant $L < μ$ :
$$ \begin{align} \Pr[X\leq L] &= \frac12 + \frac12\operatorname{erf}\frac{L-\mu}{\sqrt{2}\sigma} \\ &\approx A \exp \left(-B \left(\frac{L-\mu}{\sigma}\right)^2\right) \end{align} $$ https://en.wikipedia.org/wiki/Error_function
Moreover i would like to know if I can use it for the following statement : Let $X_k$ iid $N(0,1)$ random variables for $k = 1...n$, $M_n = \underset{k=1...n}{max} X_k$
Then $Pr[M_n \leq L] \approx A^n e^{\left( -nBL^2\right)}$
In terms of approximation, I do not see how this form could be obtained.
Using $$\Phi(x)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{t^2}{2}}\,dt=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ Plotting, what you could notice is that the function $$f(x)=\frac{ e^{-\frac{x^2}{2}}}{1-\Phi(x)}-2$$ is "close" to linearity.
Generating data for $0\leq x \leq 5$ and making the numbers rational, a quick and dirty regression gives (with $R^2=0.99999825$) $$f(x) \sim \frac{1356 }{743}x^{1194/1069}$$ and, whatever $f(x)$ could be, $$\Phi_{\text{app}}(x)=1-\frac{e^{-\frac{x^2}{2}}}{f(x)+2}$$
Just to give an idea $$\int_0^\infty \left(\Phi(x)-\Phi_{\text{app}}(x)\right)^2\,dx=8.33\times 10^{-6}$$