Sulfur trioxide violates the octet rule. Upon drawing the Lewis dot structure for sulfur trioxide, we see that the central sulfur atom is bonded to three other oxygen atoms by double covalent bonds. Therefore, sulfur is in fact surrouded by 12 electrons and not 8 as per the octet rule. This is explained by the fact that the excited state of sulfur has the following electronic configuration: $1s^22s^22p^63s^13p^34s^03d^2$. As there are 3 oxygen atoms, $sp^2$ hybridised orbitals take part in single $\sigma$ covalent bonds with the unpaired $p_y$ orbitals of the oxygen atoms. In sulfur, there still exist one unpaired $p$ orbital and 2 unpaired $d$ orbitals. These remaining unpaired orbitals take part in $\pi$ covalent bonds with the unpaired $p_z$ orbitals of the oxygen atoms. My question is, why do the electrons in sulfur from the $3s$ and $3p$ orbitals in the ground state promote themselves to the $d$ orbitals rather than one electron going into the intermediate $4s$ orbital and the next electron going into the first $3d$ orbital? After all, according to the Aufbau Principle, $4s$ is less energetic when compared to $3d$.
I feel you are confused on atomic orbital and molecular orbital.
In the excited state of a single sulfur atom, electron do fill 4s first. You are right here. However, it is a different story in molecules.
In sulfur trioxide the bonding electrons are filling the molecular orbital, not the atomic orbital such as 4s or 3d. It is forbidden that both 3s and 4s participate any hybridization or formation of molecular orbital. So no matter how low is the energy of 4s, because 3s is already participating, 4s can not be part of the bonding.