Deformed atom vs deformed nucleus I am not an expert in nuclear physics, but according to my impression, many nuclei are deformed, i.e., they are not spherical but ellipsoidal.
Could the electrons in an atom also have a deformed distribution?
 A: An important difference between atoms and nuclei is that the interaction between electrons is long-range and repulsive, while the interaction between nucleons is short-range and attractive. This means that nucleons want to maximize their spatial overlap. This can be achieved by preferentially filling a particular $m$-state of several orbits, rather than all $m$-states of a single orbit. (See pictures in anna v’s answer). This tendency is counterbalanced by the difference in energy of the levels, so whether or not there is deformation depends on the single particle spacing near the fermi surface. Large gaps lead to spherical shapes, while a high level density leads to deformation.
In contrast, electrons in an atom want to minimize overlap, so they tend to fill different $m$ states of the same $l$ orbit, yielding roughly spherical shapes.
A: This depends somewhat on what you mean by deformation.
Except for s orbitals, the atomic orbitals lack spherical symmetry. (This is the opposite of what is claimed in the incorrect answer by anna v.) So for example, boron has all its electrons in s states except for one in a p state. The ground state of the system is therefore three-fold degenerate. Within this three-dimensional space, you can pick a state, for example, that has a nonvanishing electric quadrupole moment $Q_{zz}$ along the $z$ axis. So in this sense, the ground state of a boron atom can be deformed.
But when we talk about deformation in nuclear physics, we mean something very different from this -- the standard is higher. What we mean is that there is a collective symmetry breaking in which many nucleons are all correlated, not just one. There are two standard experimental signatures of this: (1) static or transition E2 moments that are much bigger than the single-particle values, and (2) a J(J+1) spacing of energy levels. These criteria are never met by atoms (at least in low-energy states that can be easily prepared), although they are often met by molecules.
The question asks yes/no, not why, but ragnar's answer does a pretty good job of explaining the why. In addition to the reasons listed by ragnar, I would also point out that the mean field in an atom is half due to the electrons but half due to the nucleus. The nucleus's field is immune to polarization.
The answer by anna v says:

The strong force is responsible for the "shape" of the nucleus in an
interplay with the electromagnetic repulsion of protons and the
mediating existence of neutrons.

This explanation, which doesn't invoke any quantum mechanics, is wrong. Deformation exists even in very light nuclei for which electromagnetic interactions (which scale like $Z^2$) are negligible. Deformation in nuclei is a purely quantum-mechanical effect. It arises when the density of single-particle states near the Fermi level is lower for the deformed field than for the spherical field. Coulomb repulsion in heavy nuclei does help to support deformation, but is not necessary -- deformation occurs without it as well.
A: The short answer is no,  the orbitals of the electrons have spherical symmetry .


the relative size and shape of atoms (top) and AIM (bottom). The atoms, from left to right, are H, F, Cl, Li, C, N, O, and F, .....

The reason this happens is because the nucleus is mainly held together by the strong interaction, whereas the electrons have orbitals about the nucleus mainly with the electromagnetic interaction.
The strong force is responsible for the "shape" of the nucleus in an interplay with  the electromagnetic repulsion of protons and the mediating existence of neutrons.
