$\require{\cancel}$
...For large samples, as a consequence of the central limit theorem, the likelihood function approaches a gaussian, whose expected value is equal to the maximum likelihood estimate. Indeed, expanding the logarithm of the likelihood function around its maximum,
$$\ln{\mathscr{L}(\theta)}=\ln{\mathscr{L}}(\hat\theta)-\dfrac{1}{2}\left|\dfrac{\partial^2\ln\mathscr{L}}{\partial {\theta}^2}\right|_\hat\theta\left(\theta-\hat\theta\right)^2 + \dots$$
for a "sufficient large" number of samples, the so called parabolic approximation can be considered, such that the likelihood function can be approximated by a gaussian-like expression, i.e.,
$$\mathscr{L}(\theta)\propto e^{-\dfrac{1}{2}\left|\dfrac{\partial^2\ln\mathscr{L}}{\partial {\theta}^2}\right|_\hat\theta(\theta-\hat\theta)^2}...$$
I tried the Taylor expansion at $ \hat\theta$ as:
$$\ln{\mathscr{L}(\theta)}= \ln{\mathscr{L}}(\hat\theta) + \cancel{\left. \dfrac{\partial \ln\mathscr{L}}{\partial \theta}\right|_{\hat\theta}\left(\theta-\hat\theta\right)} \color{red}+\dfrac{1}{2}\left. \dfrac{\partial^2\ln\mathscr{L}}{\partial {\theta}^2}\right|_\hat\theta\left(\theta-\hat\theta\right)^2 + \dots$$
Why the minus sign?
The minus sign comes from the fact that at the maximum of ln(L), the second order condition requires that necessarily ln(L)''<=0. And the opposite of the absolute value of ln(L)'' is equal in this case to ln(L)''.