Here is an explanation that, I believe, solves most of the problems raised by others on this thread.
TL;DR
When we bring the atoms closer from infinity, electrostatic repulsions come into play between the electrons of the two atoms (as <PERSON> out under <PERSON>'s answer, the energy of the orbitals has nothing to do with nuclear repulsion). These inter-electronic repulsions increase the potential energy of the system.
Since the energy increases, the average energy of the bonding and anti-bonding molecular orbitals also increases. Thus, the bonding molecular is not lowered in energy (on the molecular orbital diagram) to the extent expected, and the anti-bonding molecular orbital is raised in energy to a higher extent than expected. That is why "antibonding is more antibonding than bonding is bonding."
(Somewhat) Mathematical Explanation
For simplicity, let us assume a symmetrical, diatomic molecule $X_2$ (we do not know how many electrons are present in it). Let us also consider only the $1s$ atomic and molecular orbitals.
When we talk of the energies of atomic orbitals, we talk of atoms ($X$'s) at infinite separation. Let the energy of the $1s$ atomic orbital of $X$ be $E_X$. Then the total energy of the $1s$ atomic orbitals in two $X$ atoms is
$$E_i = 2E_X$$
However, when we talk of the energies of molecular orbitals, we must remember that the atoms are now a finite distance away. Let us consider the bonding and anti-bonding molecular orbitals separately.
Let the ideal/expected energy (without taking into account the inter-electronic repulsion) of the $\sigma 1s$ molecular orbital be $E_1$. 724
However, there will be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms. Let the extra energy due to repulsions be $E_{e_1}$. Then the energy of the $\sigma 1s$ molecular orbital will be
$$E_\sigma = E_1 + E_{e_1}$$
Again, let the ideal/expected energy of the $σ^ 1s$ molecular orbital be $E_2$. However, there will still be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms (we do not need to know whether the repulsions are greater in the bonding or anti-bonding molecular orbital; just that they exist). Let the extra energy due to repulsions be $E_{e_2}$. Then the energy of the $σ^ 1s$ molecular orbital will be
$$E_{\sigma ^*} = E_2 + E_{e_2}$$
Therefore, the total energy of the $1s$ molecular orbitals is
$$E_f = E_\sigma + E_{\sigma ^*} = (E_1 + E_{e_1}) + (E_2 + E_{e_2}) = (E_1 + E_2) + (E_{e_1} + E_{e_2})$$
The term $(E_{e_1} + E_{e_2})$ arises due to the change in distance between atoms, i.e. from infinite to finite separation. Therefore it is the change in potential energy of the system, or the work done in bringing the two atoms to the final finite distance.
$$W = E_{e_1} + E_{e_2}$$
Now, we can apply the law of conservation of energy:
$$E_i + W = E_f$$ $$2E_X + (E_{e_1} + E_{e_2}) = E_\sigma + E_{\sigma ^}$$ $$E_\sigma + E_{\sigma ^} - 2E_X = E_{e_1} + E_{e_2} > 0$$ $$E_\sigma + E_{\sigma ^} > 2E_X$$ $$\frac{E_\sigma + E_{\sigma ^}}2 > E_X$$
Therefore, the average energy of the bonding and anti-bonding molecular orbitals is greater than that of the atomic orbitals.
To put it another way,
$$E_\sigma + E_{\sigma ^} > 2E_X$$ $$E_{\sigma ^} - E_X = E_X - E_\sigma$$
Therefore, "antibonding is more antibonding than bonding is bonding. 374