The real locus of the moduli space of stable curves of genus zero with $6$ marked points is a non-formal space with vanishing Massey products.
This is a compact non-orientable 3-manifold that is hyperbolic on the complement of $10 = \frac{1}{2} { 6 \choose 3 }$ embedded tori.
See the remark on page 4 of my paper with Etingof, Kamnitzer, and Rains.
The rational cohomology algebra of that manifold is generated by symbols
$\nu_{ijk}$ for $1\le i\lt j\lt k\le 5$, and has defining relations given by
$$
\nu_{ijk}\nu_{ijl}=0.
$$
See Proposition 2.3 of the above paper (the second relation in loc. cit. doesn't occur because $6$ is too small).
The Massey products vanish because their indeterminacy is too big: there is no room for them to be non-zero..... wait, I need to check this.
I forget if there is a good argument about why the Massey products should vanish.
But I redid the computation $\langle \nu_{123}, \nu_{234}, \nu_{345}\rangle$ and found zero.
The way I did the computation was by representing the cohomology classes by explicit submanifolds, and doing intersections.
More precisely:
• $\nu_{123}$ is represented by the submanifold $M_{123}$ where the marked points "0" and "1" are separated from the marked points "2" and "3" by a normal crossing.
• $\nu_{234}$ is represented by the submanifold $M_{234}$ where the marked points "0" and "2" are separated from the marked points "3" and "4" by a normal crossing.
• $\nu_{345}$ is represented by the submanifold $M_{345}$ where the marked points "0" and "3" are separated from the marked points "4" and "5" by a normal crossing.
At this point, one needs to check that the manifolds $M_{123}$ and $M_{234}$ are transverse, and that the manifolds
$M_{234}$ and $M_{345}$ are transverse (a necessary condition for this kind of computation to work out).
• The intersection $M_{123}\cap M_{234}$ is bounded by $\frac {1} {2}$ times the manifold where the points "0", "1", "2" are separated from the points "3", "4".
Call that manifold $A$.
• The intersection $M_{234}\cap M_{345}$ is bounded by $\frac {1} {2}$ times the manifold where the points "0", "2" are separated from the points "3", "4", "5".
Call that manifold $B$.
Finally, we need to compute the cohomology class of $X := (M_{123}\cap B) \cup (A \cap M_{234})$. That manifold represents the Massey product $\langle \nu_{123}, \nu_{234}, \nu_{345}\rangle$. Unfortunately, $X$ is not empty. But you can intersect it with all the generators of $H_2$ and see that you always get zero. The generators of H_2 are the tori I talked about in the beginning of my answer. The manifold $X$ is actually disjoint from all but one of these tor, and intersects that last torus non-transversely. Even worse: it is contained in that torus. One then needs to deform it and see that it can be made disjoint from that last torus.
I guess I should also check the Massey product $\langle \nu_{123}, \nu_{234}, \nu_{123}\rangle$...
Griffiths and Morgan wrote a fine book on the subject. Apart from the obvious attractiveness of learning a theory from its creator, it is written in an amazingly user-friendly style. For example, Chapter XIII is devoted to examples and computations: it starts with the computation of a minimal model for the forms on a sphere
and ends with Massey triple products on compact Kähler manifolds, a section inspired by the 1974 Inventiones article of Deligne,Griffiths, Morgan, Sullivan. The first hundred pages (Chapters I to VII) are an introduction to the necessary algebraic topology and you can probably essentially skip it, judging from your description of what you already know.
Reference Griffiths, P.; Morgan, J. (1981), Rational homotopy theory and differential forms, Progress in Mathematics, 16, Birkhäuser
Best Answer
In this paper by Tornike Kadeishvili, it is shown that the rational cohomology of a simply connected space carries the structure of a $C_\infty$-algebra, and that the isomorphism type of this object determines the rational homotopy type of the space.