I don't understand how the 'cyclopropenium' ion is aromatic. According to my understanding, the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital - since each carbon is bonded to 3 other atoms and has 4 valence electrons. How does that lead to aromaticity, since this doesn't fullfil Huckel's rule?
It does meet Huckel's rule. When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons. Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.
A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms. Thus aromaticity is especially powerful in a cyclopropenyl cation ring. This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.
the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital
This is incorrect.
The requirements for Huckel aromaticity are:
There must be a continuous, planer ring of overlapping $p$ orbitals.
There must be $4n+2$ $\pi$ electrons in the system.
The cyclopropenyl fulfills these requirements because it is planar, has overlapping $p$ orbitals, and has 2 $\pi$ electrons ($n = 0$).
The cyclopropenyl cation is aromatic because it is meeting all the definitions of Huckel's rule of aromaticity:
- All carbons are sp2 hybridised.
- There are 3 carbon atoms which form a conjugated system and moreover it has a positive charge therefore it has $3$-$1$=$2$ pi electrons.
- If we put $n=0$ in Huckel's expression ($4n+2$) pi electrons we are getting 2 as the result.
- The cation also has overlapping planar and parallel p orbitals. You may be knowing that for maximum overlapping, orbitals should be parallel to one another and planar.
Since the required criteria are fulfilled, it is aromatic.