The subtlety is that an arbitrary wavefunction doesn't reduce to a point of the classical phase space in the limit $\hbar \to 0$ (thinking about phase space makes more sense since in the classical limit one should have definite coordinates and momenta).
So one could ask, which wavefunctions do. And the answer is that the classical limit is built on the so-called coherent states -- the states that minimize the uncertainty relation (though I don't know any mathematical theorem proving that it's always true in the general case, but in all known examples it is indeed so). States close to the coherent ones can be thought of as some "quantum fuzz", corresponding to the quasiclassical corrections of higher orders in $\hbar$.
Example of this for the harmonic oscillator can be found in Landau Lifshits.
Regarding the fluid argument. About your remark (1): the $|\psi|^2$ for the stationary state is indeed stationary, but it still satisfies the continuity equation since the current is zero for such states. Your remarks (2) and (3) are quite right because, as I already said, the classical limit can't be sensibly taken for arbitrary states, it is built from coherent states.
And also I must admit that the given fluid argument indeed doesn't provide any classical-limit manifestation. It's just an illustration that "everything behaves reasonably well" to convince readers that everything is OK and to presumably drive their attention away from the hard and subtle point -- it often happens in $\it{physics}$ books, probably unintentionally :). The problem of a nice classical limit description is actually an open one (though often underestimated), leading to rather deep questions, like the systematic way to obtain the symplectic geometry from the classical limit. In my opinion it is also connected to the problem of quantum reduction (known also as the "wave function collapse").
I'm answering only part of your question, please bear in mind that some of the topics that you pointed out are doubts I have myself.
Concerning the comment from Feynman, I usually don't like the authority arguments. If your only reason to believe something is that someone said it, it's not a good reason. Just as some examples, Newton believed that light had no wave-like behaviour, which is simply wrong.
About the variational formulation of Schrödinger Equation. If by variational you mean, having a Lagrangian, the following Lagrangian does the job:
$
\mathcal L = \frac{i\hbar}{2}\left(\psi^*\partial_t \psi - \psi \partial_t \psi^*\right) - \frac{\hbar^2}{2m}\left(\nabla \psi\right)\cdot \left(\nabla \psi^*\right) - V\psi^*\psi
$
With the assumption that you have a complex classical scalar field, $\psi$, and that you can calculate the Euler-Lagrange equation separately for $\psi$ and $\psi^*$. As with anything using the extremum action principle, you really have to guess which your Lagrangian is based in Symmetry principles + some guideline for your problem, here is no different.
I don' t know how did Schrödinger made the original derivation, but if you make a slight change of variables in the above lagrangian, you get very close to what you have written above.
The trick is writing $\psi = \sqrt n e^{iS/\hbar}$, where $n$ is the probability density and $S$ is, essentially the phase of the wave-function. If you have trouble with the derivation, I can help you latter.
Again about Feynman's quote. It's possible to arrive at Schrödinger equation without passing through an arbitrary heuristic procedure, that way is developed by Ballentine's book. You still have to postulate where you live, if you consider that arbitrary or not, it is completely up to you.
The other point that it is possible to consistently and rigorously construct a quantum theory from any classical mechanical theory, using quantization by deformation. That's why I think it's a bit false that "It's not possible to derive it from anything you know. It came out of the mind of Schrödinger", as Feynman, and many other great names, said.
Best Answer
I am confused by the comments. Perhaps I am missing something.
For any linear differential equation with a given solution, a constant times that solution is also a valid solution.
The Schroedinger equation is a linear differential equation for the wavefunction $\Psi$. This means that if $\Psi$ is a solution then so is $A\Psi$ where $A$ is a complex constant. You can fix this constant by further requirements like the normalization condition $\int d^dx |\Psi|^2=1$. Of course as Peter Morgan said, the phase of $A$ remains undetermined by the normalization condition.
I think this is also what Griffiths means. I quote him after he says that the wavefunction should be normalized since the particle has to be somewhere.
and then he goes on to say that we must pick $A$ such that the wavefunction is normalized.