As I recall, the Cayley projective plane is painful to build, but it is a 2-cell complex, with an 8-cell and a 16-cell. The cohomology is Z[x]/(x^3) where x has degree 8, as you would expect. Its homotopy is unapproachable, because it is just two spheres stuck together, so you would pretty much have to know the homotopy groups of the spheres to know it. The attaching map of the 16-cell is a map of Hopf invariant one, from S^15 to S^8, the last such element.
I think the real reason that the Cayley projective plane exists is because any subalgebra of the octonions that is generated by 2 elements is associative. That is just enough associativity to construct the projective plane, but not enough to construct projective 3-space. And this is why you should not expect there to be a projective plane for the sedonions (the 16-dimensional algebra that is to the octonions what the octonions are to the quaternions), because every time you do the doubling construction you lose more, and in particular it is no longer true that every subalgebra of the sedonions that is generated by 2 elements is associative.
Mark
$\DeclareMathOperator\Tr{Tr}
$Suppose that $k$ is a field with $\operatorname{char}(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.
When $x \in A \setminus k$, the trace $\Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure: $x$ is the root of a quadratic polynomial with coefficients $-\Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not; it doesn't really matter.
It is an old theorem (Jacobson? Albert? I can't recall … check "The Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley–Dickson doubling process, the norm form $N$ is a Pfister form, i.e., $N$ is isomorphic to $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ for some $a,b,c \in k$.
It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $\langle1,-1\rangle \otimes \langle1,-1\rangle \otimes \langle1,-1\rangle$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.
In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ depends only on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$ whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.
Best Answer
Yes.
As referenced above, G=PSL(2,7) has 168 elements and is the automorphism group of the Klein quartic as well as the symmetry group of the Fano plane.
There are 30 possible ways of arranging 7 objects into 7 triads such that each pair of objects belongs to exactly 1 triad. There are 16 possible 7-bit sign reversals used to generate the 30*16=480 valid Octonions (of which Baez' shows one). All 480 permutations of Fano plane and cubic are shown here (15MB pdf).
There are 168 permutations of "twisted Octonions" for each of the 30 sets of triads (168*30=5040=7!).
The reference of Baez' to the twisted Octonions are treated in more detail by Chesley here. They have been integrated in my work with Mathematica code based on his c code. It was used to produce the WikiPedia referenced Fano plane and cubic in the posted question. I have also validated the 7! twisted Octonion permutations with it.
The Baez Fano cubic is shown in a pic here (sorry, I can't post pics in MathOverFlow yet). It has yellow nodes highlighting the Real, Complex, Quaternion and Octonion plane. It is one of the few Octonions that have the sign mask (per Chesley) of 00, meaning it is one of the canonical 30 sets of 7 triads.
My term "flipped" (in the pic referenced above and the .pdf files) refers to the need to rearrange the Fano mnemonic nodes from the algorithms default "flattened" sequence in order to apply the same directed arrow logic as the "non-flipped" Fano planes and cubics. The flattening procedure is simply taking the first occurrence of each number of 1-7.
The Octonions are known to be associated with the 240 vertices of E8. It is the relationship between the flipped and non-flipped Octonions that provides the 2 to 1 mapping to E8. These are shown integrated in my (as yet) very speculative work on a ToE, shown in a comprehensive (40MB pdf) here.