Does a free electron, one that's not either in an atom or a wire, have an associated wave-function? Would a free electron, one that's not either in an atom or moving through a wire, but moving through empty space on its own, have an associated wave-function? Or, is an electron described as a wave-function only when inside an atom?
And if the type of free electron I'm describing does have a wave-function, how can it when the electron wouldn't be in an orbital within an atom?
 A: Yes, a free particle has a wavefunction. If it has sharp momentum $p$, it is given by the plane wave
$$ \psi_p(x) = \mathrm{e}^{\mathrm{i}px}$$
which might look a bit strange, because it is non-normalizable/not square-integrable, but that should not be all too troubling - no momentum uncertainty at all is rather unphysical, after all. The general wavefunction of a free particle is a square-integrable superposition of these plane waves. As to why such non-normalizable states arise, read Rod Vance's answer on rigged Hilbert spaces and their connection to scattering states.
Electrons are always quantum objects, whether they are bound or not, and thus must always be represented as a quantum state, which is almost always representable by a wavefunction.
A: There is no need of a potential for the Schrödinger equation to have a solution, namely
$$
i\hbar \frac{\partial}{\partial t} |\psi\rangle = H |\psi\rangle
$$
does possess solutions even when the Hamiltonian contains no potential. Questions of normalisability may arise, but that is another point (and can anyway be solved by expanding in Fourier terms).    

And if the type of free electron I'm describing does have a wave-function,   how can it when the electron wouldn't be in an orbital within an atom?

You seem to have a little odd idea of what a wave function is, probably due to some odd-written books in chemistry which associate the wave function to some sort of orbital cloud. That is by no means what the wave function is. Rather it is the representation on the position basis  $\langle x|\psi\rangle$ of the solution of the Schrödinger equation above.
