The test involves a partial sum of a diagonal whose second element is a prime. Because of the symmetry of Pascal Triangle each prime has two diagonals parallel to the sides of the triangle. These two diagonals meet at the central binomial coefficient (cbc). The number of elements of each diagonal up to the cbc is always $(p-1)$ (with of course the cbc element included).
Observation 1: every element (except the first one equal to $1$) of such a diagonal is divisible by the prime $p$. This property is not shared by composites which have only some elements divisible by the composite.
Because of such divisibility property, it makes more sense to add all the elements up to and including the cbc before testing if the sum is divisible by the given prime. Therefore the sum must be calculated. It turned out that the next element after the cbc, the $p-th$ element, provides that sum after subtracting $1$ (the first element of a diagonal). This comes from the fact that the $p-th$ element and the elements not aligned with it form a hockey stick. So the sum of $(p-1)$ elements is given by the $((p-th) -1)$.
Observation 2: For a prime, the sum of $(p-1)$ elements is always divisible by $p^3$. No proof is avalaible at this time.
So there is a need to calculate the sum using the hockey stick value which itself needs to be calculated. It turns out that the hockey stick value is half of the $p-th$ cbc. The cbc value is given by $v=(2n)!/(n!)^2$.
Here's a numerical example to illustrate the method.
We consider $p=5$. The $(5-1)=4$ elements are $5,15,35,70$. The value of $n$ is always given by the value of $p$ so $n=5$ and $v=10!/(5!)^2=252$. The value of the sum is simply $s=(252/2 - 1)=125=5^3$. $5$ is the only prime for which the sum is $p=5^3$.
Other sums involving larger primes than $5$ will have other factors beside $p^3$.
For $p=7$, we have $s=(3432/2 -1)=1715=5*7^3$
For $p=11$, we have $s=(705432/2-1)=352715=5*53*11^3$
There is a table of $500$ values of central binomial coefficients here:
https://oeis.org/A000984/b000984.txt
Question: How does this primality test compare to other well-known methods?
Your "Observation 2" is Wolstenholme's theorem. The converse is still an open problem, and regarded as quite hard.
On the time complexity of the algorithm, it's pretty bad. It requires calculating $\binom{2n-1}{n-1} \mod n^3$, which will take $\mathcal{\Omega}(n)$ time. It will probably be quite a bit worse, as the numbers you calculate will be massive, and it's not obvious how to avoid using big integers and keep computations modulo $n^3$. For comparison, the simple trial divison algorithm runs in $\mathcal{O}(\sqrt{n})$.