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.
As mentioned in the comment, such a transpose map on $W$ and $V$ is coordinate dependent.
If you want a coordinate independent map, you'll have to stick with the map from $W^*$ to $V^*$.
For a finite dimensional vector space, the double dual space of $V$ (i.e. $V^{**}$) is naturally isomorphic to $V$ (natural here has a technical definition, but essentially means this isomorphism does not depend on coordinates).
On the other hand, even though the dual space of $V$ (i.e. $V^*$) and $V$ are isomorphic, we do not have that they are naturally isomorphic. Picking an isomorphism is nearly the same thing as selecting an inner product (i.e. adding geometry to $V$).
Given a linear map, $A:V \to W$, we can define $A^T:W^* \to V^*$ naturally via $A^T(f)(v)=f(A(v))$. [$f:W \to \mathbb{R}$ is a linear functional on $W$, $A^T(f):V \to \mathbb{R}$ is a linear functional on $V$.]
If we want to define $A^T$ as a map from $W$ to $V$, we'll need to pass (somehow) from $W^*$ to $W$ and $V^*$ to $V$. One way to do this is via inner products.
Suppose that $\langle v_1,v_2 \rangle_V$ is an inner product on $V$ and $\langle w_1,w_2 \rangle_W$ is an inner product on $W$. Then $v \mapsto \langle v, \cdot \rangle_V$ gives an isomorphism from $V$ to $V^*$. Let's give this a name: $\varphi_V(v)=\langle v, \cdot \rangle_V$. Likewise, $w \mapsto \langle w,\cdot \rangle_W$ from $W$ to $W^*$ denoted $\varphi_W(w)=\langle w,\cdot \rangle_W$.
Then we have: $A^T(\varphi_W(w))(v) = (\varphi_W(w))(A(v)) = \langle w, A(v) \rangle_W$. If we then use $\varphi_V^{-1}$ we can turn the linear map $A^T(\varphi_W(w))$ into a vector in $V$. Thus composing maps as follows: $\varphi_V^{-1} \circ A^T \circ \varphi_W$ gives us your desired transpose map:
$$\varphi_V^{-1} \circ A^T \circ \varphi_W:W \to V$$
If we call this $\widetilde{A^T} = \varphi_V^{-1} \circ A^T \circ \varphi_W$, we have
$$ \langle \widetilde{A^T}(w),v \rangle_V = \langle w, A(v) \rangle_W$$
for all $v \in V$ and $w \in W$. Some texts would just use the above equality as the definition of the transpose map (but, of course, this depends on those pesky inner products).
The issue is that turning the transpose map into a map on $W$ and $V$ requires us to pick isomorphisms between $V$ and its dual as well as $W$ and its dual. A special case of this is selecting an inner product.
Picking isomorphisms between vector spaces and their duals is very closely related to selecting bases [Given a basis for $V$, you get a dual basis for $V^*$ and bam you have an isomorphism.]
Likewise, picking an inner product is closely related to selecting a basis [Pick a basis, declare it to be orthonormal, and bam you have an inner product.]
In the end, you really can't get the transpose map to be "coordinate free" on the original vector spaces in the way you're looking for. It all comes down to the fact that a vector space and its dual are not naturally isomorphic.
Best Answer
The concept of Hermitian linear transformations requires your complex vector space to have an additional structure, a Hermitian product, i.e. a conjugated-symmetric inner product: $x\cdot y = (y\cdot x)^*$, with $^*$ denoting complex cojugation. A linear operator $A$ is then called Hermitian if $x\cdot Ay =(y\cdot Ax)^*$. The matrix of a Hermitian operator is a Hermitian matrix (in the sense that the matrix is equal to its transpose complex conjugate) IF the matrix is written in an orthonormal basis (for a general basis the matrix of a Hermitian operator is NOT in general a Hermitian matrix).