The determinant of the adjoint of a matrix $A$ is given by $|A|^{n-1}$ where $n$ is the order of the square matrix.
So, for odd $n$, the determinant is always positive.
Is there an intuitive explanation for this? It is surprising that every odd ordered matrix irrespective of its elements would have an adjoint with a positive determinant.
Here is a way to understand this in $\mathbb{R}^3$. Suppose the rows of your original matrix $A$ are coordinates of vectors $a,b,c$ in some orthonormal basis. The rows of the adjoint matrix are the coordinates of the vectors $b\times c$, $c\times a$ and $a\times b$ (or even better, the coordinates of the corresponding bivectors). The determinant of the adjoint matrix is thus the oriented volume of the parallelepiped defined by those cross-products.
We can assume that $a,b,c$ are linearly independent, otherwise at least two of the cross-products will be parallel an the adjoint determinant zero. Now observe that if $\{a,b,c\}$ is a positively oriented basis, these cross products also form a positively oriented basis (think of $i$, $j$, $k$) and hence the determinant is positive. If, however,$\{a,b,c\}$ is a negatively oriented basis, then the cross products belong to the cone, polar to the cone generated by $\{a,b,c\}$ (instead of the dual cone they belong in the former case). They are thus opposite to vectors with negative orientation (preserving their order) hence they form a positively oriented basis again. In the general multidimensional case I think you can use geometric algebra to prove it. A proof using geometric algebra will necessarily convey some geometric intuition.