How does a free electron look like? In a simple atom say hydrogen, there is an electron cloud which is spherical in shape. What about a free electron, how big or small will that cloud be? I think the term cloud here means the likelihood the electron can be found, as for a free electron in absent of other forces does it stay as a point-like particle or a standing wave spreading out across the universe?
 A: "What about a free electron, how big or small will that cloud be?"
As big or as small as you want it to be, in principle – that is, in an infinite vacuum at zero Kelvin. The most common example of a normalized wave packet is the Gaussian (because it's rotationally symmetric and computationally simple):
\begin{align}
  \psi_G(\mathbf{x}) &= \frac{1}{(2\pi)^{3/4} \sigma^{3/2}} \exp\left(-\frac{[\mathbf{x}-\mathbf{x}_0]^2}{4\sigma^2} + i\frac{\mathbf{p}_0\cdot\mathbf{x}}{\hbar}\right),
\end{align}
where the packet is centered at $\mathbf{x}_0$ and has mean momentum $\mathbf{p}_0$ with real space "width" (standard deviation) $\sigma$. $\sigma$ can, in principle, be as large or as small as you like. The only question is whether you have the energy needed to make it small because the mean energy in the electron is given by:
$$\langle \psi_G| H_{\mathrm{free}} |\psi_G\rangle = \frac{\mathbf{p}_0^2}{2m} + \frac{\hbar^2}{8m\sigma^2},$$
which diverges as $\sigma \rightarrow 0$.
Pick any form for the electron wave packet you like, and you'll get similar results. Just be careful to keep the packet normalizable, otherwise you'll end up with annoyances caused by non-physical states like the plane waves $\psi_{\mathrm{pw}} = \frac{1}{\sqrt{2\pi}} \exp\left(i \frac{\mathbf{p}_0\cdot \mathbf{x}}{\hbar}\right)$.
A: The surrounding cloud is just a picture of a zone where the electron has a very high probability of being there. Note that we are dealing with probabilities, so the electron has the possibility (a very small one) of being in the other side of the universe. Questions like "where is exactly the electron?" or "what is the shape and size of an electron?" are, in a certain sense, "forbidden" by the Heisenberg's uncertainity principle, it doesn't matter if the electron is bounden to an atom or if it's a free electron.
A: Since it's isolated (and therefore has no potential energy), the quick answer is that it resembles (but is not exactly) a Gaussian distribution on the order of 10 - 50 angstroms wide. You can calculate it yourself from the equation for thermal wavelength:
$$
\Lambda = \sqrt{\frac{h^2}{2\pi m k_BT}} 
$$
At room temperature, using the mass of an electron we have $\Lambda = 4.3\times10^{-9}$ meters, or 43 Angstroms.
So it would "look" something like this, where red indicates a "denser" part of the "cloud" (though this 2D gaussian doesn't do justice to a 3D function):

This is a 3D gaussian. We have to use cross sections to see the probability density though:

