Well, one trivial connection is that if you look at $1\times 1$ matrices (which have only a single complex entry), then you'll find that it is real iff it is Hermitian, its complex conjugate is its conjugate transpose, and its polar decomposition is the polar form.
Also, just like a complex number can be uniquely decomposed into a real and an imaginary part ($z = a+\mathrm ib$ with real $a,b$), a complex matrix can be uniquely decomposed into a Hermitian and an "anti-Hermitian" part, i.e,. $M =A + \mathrm iB$ with $A$ and $B$ Hermitian. And just like $\Re(z)=\frac12(z+\bar z)$ and $\Im(z)=\frac1{2\mathrm i}(z-\bar z)$, the Hermitian part of a matrix is $\frac12(M+M^*)$ and the "anti-Hermitian" part is $\frac1{2\mathrm i}(M-M^*)$.
Moreover, just like $\bar zz$ is a non-negative real number, $M^*M$ is a positive semidefinite matrix.
Another point: Hermitian matrices have real eigenvalues, and unitary matrices have eigenvalues of the form $\mathrm e^{\mathrm i\phi}$.
About the usefulness of the analogy:
In classical physics, observables should be real. In quantum physics, observables are represented by Hermitian matrices. Also, the quantum analogue to probability densities, which are non-negative functions with integral $1$, are density operators, which are positive semidefinite matrices with trace $1$. So there's indeed some connection.
Since complex conjugation satisfies $\overline{xy} = \overline{x} \cdot \overline{y}$ and $\overline{x+y} = \overline{x} + \overline{y}$, you can see with the Leibniz formula quickly that $\det[A^*] = \overline{\det[A]}$.
For complex matrices $\det[A] = \det[A^T]$ still holds and doesn't require any changes to the proof for real matrices.
Together this means that $\det[A] = \overline{\det[A^H]}$.
This applies to the eigenvalues as well: the characteristic polynomial of $A^*$ is given by $\det[tI - A^*] = \det[(\overline{t}I - A)^*] = \overline{\det[\overline{t}I - A]}$ and the eigenvalues of $A^*$ are exactly the complex conjugates of those of $A$.
In particular if $A$ is hermitian, $A = A^*$ and so all eigenvalues are equal to their complex conjugates - in other words, they're real.
Best Answer
For $z,w \in \mathbb{C}$, we have $\overline{zw} = \overline{z} \overline{w}$ and $\overline{z+w} = \overline{z} + \overline{w}$ etc.
The determinant is obtained by performing various addition and and multiplication operations on its entries. Since complex conjugation can be done before or after these operations, your claim $\overline{\det A} = \det \overline{A}$ holds.
Regarding your last sentence, note also that transposing a matrix does not change its determinant.