It's definitely not a matter of available room. An electron can fit in arbitrarily small room, according to any experiment humans have been able to conduct. Nor is 1s so much smaller than the others.
But little room implies a bigger kinetic energy of the electron. More generally, a big change in the wavefunction over a short distance, for instance a change of sign, or a big fading, implies much kinetic energy; if you like Fourier transforms, the short distance needs components with a big wave number k, which contribute much kinetic energy (ħk)
2/(2m).
One consequence it that there is one optimum size for the orbital of lowest energy, 1s: a smaller orbital puts the electron nearer to the nucleus and improves the electrostatic energy, but the kinetic energy worsens brutally. You can write simplistic models for the sum of both energies and find an optimum orbital size that is numerically meaningful. An other consequence is that the 1s orbital shrinks when the nucleus contains more protons, because the stronger electrostatic attraction can afford a bigger kinetic energy.
An other consequence is that a wavefunction that changes its sign more times
in a given volume (this is simplified thinking) implies a bigger kinetic energy while the electrostatic energy isn't better, so orbitals with more nodes have a higher energy, that is, the electron is less bound to the nucleus.
So an orbital with a radial node (2s) or an azimutal node (2p) binds the electron less than without a node (1s).
Now, we call
conventionally "1" the orbital with zero nodes, "2" the orbitals with one node, and so on. So the impossibility of an azimutal node in the "1" orbital results from the naming convention, because the first such orbital is called "2".
It's the very nature of electrons, and of all fermions as opposed to bosons like the photon, that they can't be in the same state. It's a mathematical impossibility, not a repulsion force. Maybe someone else can explain this. And because electrons can have two states of spin, in 1s state you can find 2 electrons, with opposite spins. Same for 2s: 2 electrons. And same for 2p
x, 2p
y and 2p
z, which are called collectively 2p but exist with three different orientations. Two electrons in each 2s, 2p
x, 2p
y and 2p
z make 8 electrons in all "2" orbitals.
In case you wonder if an orbital could have the simple decaying electron density of 1s and at the same time the azimutal behaviour of a p orbital, the answer is no. "p" means that the wavefunction makes one phase turn (well, for some p orbitals at least) for each geometric turn around the nucleus, and since at the nucleus the wavefunction must have every phase from 0° to 360° its only possible value is zero. More generally, only the "s" orbitals have nonzero value at the nucleus.
It happens that each radial node has the same effect on the orbital's energy as each azimutal node for the hydrogen atom.
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html third frame there
That's why we add the numbers of radial and azimutal nodes and group the orbitals in 1, 2, 3, 4... Though, this is inaccurate for more electrons, and higher orbitals interleave
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.htmland this is just for one outer electron at sodium. Transition elements do it more.