While deriving the wave function why don't we take into the account of the probability density of the nucleus? My intuition says that the nucleus is also composed of subatomic particles so it will also have probability cloud like electrons have. Do we not do it for simplicity of the calculation, or is the nucleus fixed, or any other property of the nucleus?
Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as we know.)
Uncertainity principle requires that the nucleus as a whole has some spatial spread.
The easy part is that the "probabilistic cloud" of a nucleus and it's constituents is WAY smaller than the space electrons pretend to occupy. That's what makes the point approximation viable.
Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient.
Note that for the hydrogen atom this approximation is not required. In this two particle case the wave function describes the relative motion and position of electron and nucleus.
The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have
where $\Psi_p$ is the wavefunction of the proton, $\Psi_e$ is the wavefunction of the electron, and $\alpha$ is the ratio of the masses. Because $\alpha$ is large, one can often approximate the proton as being fixed at one point.
The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.
Nuclei are quantum particles and have a wavefunction and hence a probability density, too. However, it is hard to calculate and to visualize, and it is often not needed.
For electrons, you can e.g. look at a one-electron density of a many electron system or at orbitals, which correspond the best possible solutions of approximating the full many-electron problem with an independent-particle problem. However, electrons are all the same, which makes the problem easier.
In contrast, if you want to visualize the nuclei of a molecule, you have a hard time. An example where it was actually done is this article. In the article, the nuclear probability densities of molecules are obtained by approximating the nuclear wavefunction as product of harmonic oscillator functions in the normal modes and by integrating over all but the coordinates of one nucleus. You see that the spatial extend of the nuclear probability density is small even for vibrationally excited states, hence it is usually of little interest.
Another problem when you want to calculate a probability density is how to treat the invariance with respect to translation and rotation of the whole system. For electrons in a molecule this is not a problem if the nuclei are assumed to be fixed in space, but you always need some reference, otherwise the density is just a constant.