Where is {det = 0} a manifold?

1. Gradient. Expanding $\det$ along row $i$: $\det A=\sum_ja_{ij}C_{ij}$ with the cofactor $C_{ij}$ independent of the entries of row $i$. So $\dfrac{\partial\det}{\partial a_{ij}}=C_{ij}$, i.e. $\nabla\det(A)=\operatorname{adj}(A)^T$ (the cofactor matrix).

If $\operatorname{rank}A=n-1$, then $\operatorname{adj}A\ne0$: indeed $A\operatorname{adj}A=0$ and at least one $(n-1)\times(n-1)$ minor of $A$ is nonzero (rank $n-1$), which is an entry of the cofactor matrix. So $0$ is a regular value of $\det$ near $A$ and by the implicit function theorem $\Sigma$ is a smooth hypersurface near $A$.

2. The case $n=2$ at $0$. With $s=a+d$, $u=a-d$, $v=b+c$, $w=b-c$ we have $a=\frac{s+u}2$, $d=\frac{s-u}2$ and
$$ad-bc=\frac{s^2-u^2}4-\frac{v^2-w^2}4=\frac{s^2+w^2-u^2-v^2}4.$$
So $\Sigma\cong\{(s,w,u,v)\in\mathbb{R}^4:s^2+w^2=u^2+v^2\}$, a quadratic cone. Its intersection with the sphere of radius $r$ is $\{s^2+w^2=u^2+v^2=r^2/2\}$, a product of two circles, a torus $T^2$. Hence $\Sigma\setminus\{0\}\cong T^2\times(0,\infty)$ and $\Sigma$ is the open cone over $T^2$.

*$\Sigma$ is not a manifold at $0$.* Suppose $0$ had a neighbourhood in $\Sigma$ homeomorphic to $\mathbb{R}^3$. Then the local homology $H_2(\Sigma,\Sigma\setminus\{0\})$ would equal $H_2(\mathbb{R}^3,\mathbb{R}^3\setminus0)\cong\tilde H_1(S^2)=0$. For a cone $\Sigma$ over a space $L$, however,
$$H_k(\Sigma,\Sigma\setminus0)\cong\tilde H_{k-1}(L),$$
which for $L=T^2$ and $k=2$ is $H_1(T^2)=\mathbb{Z}^2\ne0$. Contradiction. (Equivalently: the punctured neighbourhoods of $0$ retract onto a torus, which is not a $2$-sphere; a punctured $3$-ball would retract onto $S^2$.)

Remark. The singular locus of $\Sigma$ is exactly $\{\operatorname{rank}\le n-2\}$, of codimension $4$ in $M_n(\mathbb{R})$. The hypersurface is smooth elsewhere, so singularities of $\{\det=0\}$ are "rare" and invisible to a transversal curve.