For a matrix $A\in SL_{n}(\Bbb R)$ is $\nabla^{n} \det(A)$ always $0$? Let $\nabla=\frac{d}{da}+\frac{d}{db}...$ denote the sum of the partial derivatives with respect to the matrix indexes. I noticed that $A:2\times2$, $\nabla \det(A)=a+d-c-b$ and that $\nabla^2 \det(A)=1+1-1-1=0$
This also looks to be true for $A:3\times3$ with $\nabla^{3}$. Is this true for all dimensions? If so, what is the reason?
 A: $\def\sgn{\operatorname{sgn}}$
Let $X$ be the generic $n$-by-$n$ matrix $(x_{ij})$  with the $x_{ij}$ variables.  Let $D = \det X = \sum_\pi \sgn(\pi) \prod_i x_{i \pi(i)}$.
Write $\partial_{ij}$ for the partial with respect to $x_{ij}$, and  $\nabla = \sum_{ij} \partial_{ij}$.   This is not the gradient but a particular directional derivative, as pointed out in the comments by loup blanc.  It appears to be what the OP had in mind.  If it is not what the OP had in mind, I will withdraw this answer.
Then $\nabla^n$ is a sum of all possible products of length $n$  of the partials $\partial_{ij}$ (i.e. mixed partials of length $n$).  But most of these, applied to $D$, are zero because $\partial_{i j} \partial_{k l} D = 0$ if $i = k$ or $j = l$ because in these cases $x_{ij}x_{kl}$ never appears as a factor of any monomial contributing to $D$. Those partials which are non-zero each occur with multiplicity $n!$.  Thus $$\nabla^n  D = n! \sum_\sigma  \prod_i \partial_{i \sigma(i)} D = n! \sum_\sigma \sum_\pi \sgn(\pi) \prod_i \partial_{i \sigma(i)}
\prod_j x_{j \pi(j)}.$$
Let's write out a typical summand for security:
$$
\sgn(\pi)(\partial_{1 \sigma(1)}\cdots \partial_{n \sigma(1n)})(x_{1 \pi(1)} \cdots x_{n \pi(n)}) = \sgn(\pi) \delta(\sigma, \pi).
$$
Hence $\nabla^n D = n! \sum_\pi \sgn(\pi) = 0$.  
