I could prove $$adj(adj(A))=|A|^{n-2}A$$ for any Non singular matrix $A$ of order $n \times n$ as follows:
we have $$adj(A)=|A|A^{-1} \tag{1}$$ Taking Inverse on both sides we get
$$(adj(A))^{-1}=\frac{1}{|A|}A \tag{2}$$
Replacing $A$ with $adj(A)$ in $(1)$ and using $(2)$ we get
$$adj(adj(A))=|adjA|(adj(A))^{-1}=\frac{|adjA|}{|A|}A=|A|^{n-2}A$$
But how can we deduce the result of $adj(adjA))$ when $A$ is Singular?
This is false. If $A$ has rank $1$ then $adj(A)=0$ and hence $$adj(adj(A))=0 \neq A$$
For example, use $$A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}$$
Added If $rank(A)=2$, then the nullspace of $A$ is one dimensional. The relation $$A (adj(A))=0$$ then implies that the columns of $adj(A)$ are in the null space, and hence they are scalar multiples of a fixed vectors. Thus $rank(adj(A))=1$ and hence $adj(adj(A)=0$.
Therefore, you can prove:
Claim: If $n \geq 3$ and $A$ is an $n \times n$ singular matrix then $$adj(adj(A))=0 .$$
Proof: If $rank(A)=n-1$ then then the nullspace of $A$ is one dimensional. By the same argument as above you get $rank(adj(A))=1$, and hence , since $n \geq 3$, $$adj(adj(A))=0 .$$
If $rank(A)<n-2$ then all the $(n-1)\times (n-1)$ minors are $0$ and hence $$adj(A)=0$$
One way is by continuity. Replace $A$ by $A+tI$. This will be nonsingular when $t$ is small and nonzero. Then let $t\to0$.
