Thomas's answer is actually pretty nice, and I upvoted it. However, it doesn't seem to have satisfied everybody, and there are some aspects of it that I think are not quite right, or don't focus on quite the right things.
Is there any hard scientific evidence that the alpha particle is tetrahedral?
The most straightforward response to this is that the notion of a tetrahedral cluster of grapes is clearly a cartoon inspired by classical intuition, and it would be absurd on the face of it to imagine that it was an accurate model of the actual quantum-mechanical system. It's not really of interest to discuss the tetrahedral cartoon in its most literal sense, because it's silly. What is at least somewhat interesting, in principle, is to ask whether the correlations among the neutrons and protons have any properties that resemble at all the kinds of correlations we would imagine from the tetrahedral cartoon.
The discussion of correlations between nucleons seems to have caused a lot of confusion in the long comment thread under Thomas's answer, so let's discuss a more straightforward example. Consider positronium in its ground state. A standard textbook treatment would start by writing down the wavefunction in separable form as something like $\Psi(x_0)\Phi(x_1)$, where $x_0$ is the vector indicating the position of the center of mass, and $x_1$ is the position of the positron relative to the electron (or relative to the c.m.). The correlations are described by the fact that $\Phi$ really tells us the wavefunction of both particles, and these correlations are perfect due to conservation of momentum. If we wish, we can completely ignore $\Psi(x_0)$, or if we care, we can let it be a state of good momentum.
But for many-body systems, this approach becomes difficult, and the classic method of attack is to instead write down a single-particle potential and populate it with particles, using occupation numbers that obey the relevant statistics. This is much more tractable for $N>2$ particles, but it has the disadvantage that the states we construct are not states of good momentum. If we apply it to positronium, then the correlations between the electron and positron are sort of there, because they both tend to live in the same region of space, but these correlations are not described exactly. There are spurious fluctuations in the total momentum, which violates conservation of momentum.
Emilio Pisanty wrote in a comment:
However, I don't have a strong enough grasp of how body-fixed frames work in QM
When we talk about body-fixed frames in nuclear physics, it's basically a way of talking about correlations between nucleons, but using a model in a specific way. Let's make an analogy with the example of the broken translational symmetry in the case of positronium.
In nuclear physics, we often violate several good symmetries at once in the same way that I described above for positronium. For a deformed rare earth nucleus, for example, we would probably use a single-particle potential with a prolate ellipsoidal shape, and we would also introduce pairing as described by the Bogoliubov approximation. The resulting many-body wavefunctions have unphysical fluctuations in momentum $\textbf{p}$, total angular momentum $J$, neutron number $N$, and proton number $Z$. For a nucleus with mass number (i.e., particle number) $A$, the relative sizes of these fluctuations all scale down with $A$, so for many heavy nuclei, for many observables, this basically produces no problems.
The ground state of an even-even nucleus such as an alpha particle is spherically symmetric in the lab frame. It has to be, because that's how angular momentum works in quantum mechanics. An even-even nucleus can be deformed in the body-fixed frame, which is what we describe, for example, in calculations using the deformed shell model. So the fact that the helium nucleus has a $0^+$ ground state really tells us nothing about whether it has a particular deformed shape such as a tetrahedron.
So when we want to tell whether a particular nucleus is deformed in its ground state, we do not get that information from its ground-state spin. We get it from other observables. If an even-nucleus is a prolate ellipsoid (which is the shape that essentially all stably deformed nuclei have), there is a rotational band built on the ground state, with spin-parity going like $0^+$, $2^+$, $4^+$, ... The energies go like $J(J+1)$. The half-life for gamma decay down this band by E2 transitions is quite short, indicating collective motion. Semi-classically, this band is interpreted as end-over-end rotation, since a quantum rotor can't rotate about an axis of symmetry. Angular momentum can be generated about the symmetry axis only by particle-hole excitations, which display none of the observational hallmarks described above.
If helium were really configured in the kind of tetrahedral configuration shown in the cartoons, then it would have some of these rotational characteristics, but not all of them. It would certainly have low-energy rotational bands built on the ground state, but we don't observe any such bands. The ground state would lack parity symmetry in the body-fixed frame, and if we were to take the cartoons totally literally, then it would also have a large electric dipole moment. This dipole moment would vanish in the true ground state (similar to the ammonia molecule, which is a classic example described, e.g., in the Feynman Lectures). However, there would be negative-parity rotational states interlaced with the positive-parity states, and there would be strong E1 transitions between these positive- and negative-parity states. We don't observe anything like this. There is evidence that a few nuclei do have reflection-asymmetric shapes, so this is not just speculative. The alpha particle's properties look nothing like the properties we would expect for the reflection-asymmetric shape.
So there is very direct observational evidence that the structure of the alpha particle is nothing like the cartoon, not even in some vaguely semi-classical way.
There are also clear theoretical reasons why we would not expect such a structure for helium. It's doubly magic, and doubly magic nuclei never have any stable deformation in their ground state.
