Significance of phase of atomic orbitals
You have quite a few questions indeed!


  What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital?


No, see this question: Physical intuition behind negative values for wave function? and the answers within.

The phase is not related to any of the physical properties of the atomic orbital. For example the probability density of the electron in the orbital is given by $|\psi|^2$. If we write $\psi_\mathrm{ps} = \mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi)$ ("ps" for phase shifted) it is easy to see that the probability density is independent of the phase chosen $x$:

$$|\psi_\mathrm{ps}|^2 = \psi_\mathrm{ps}^*\psi_\mathrm{ps} = \mathrm{e}^{-\mathrm ix}\psi^*(r,\theta,\phi)\mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi) = |\psi^2|$$

so the phase shift has not done anything. Likewise, the energies etc. are independent of any constant factor that is placed in front of the wavefunction. This is easy to show using

$$E = \frac{\langle \psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}.$$


  What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefunction.


In hydrogen the 1s and 2s orbitals are orthogonal to each other (in general all AOs are orthogonal to all other AOs). This is because the 2s orbital has an inner radial node and two regions which are phased differently (conventionally, positive + negative; however you can draw it negative + positive and the overlap would still be zero).


  How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs?


The absolute phase of the orbital has nothing to do with it, but rather the relative phase. See the linked question above.


  So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?


I'd be inclined to say that there's no real significance, except that that's how the maths works out. More fundamentally the presence of a node (in this case an angular node) means that there must be regions of different phases. The same applies to the 2s orbital which has a radial node instead of an angular node. This is analogous to the solutions to the particle in a box: the ground state solution has no nodes and is therefore positive throughout, but the first excited state has one node and therefore has one positive and one negative region.