I'll give a shot at an answer. The relevant dimensions are of the form $2^j-2$. For
$j\leq 4$, it is easy and classical that we can construct manifolds of Kervaire invariant one. The
problem was ``reduced'' from differential topology to pure stable homotopy
theory by Browder in 1969. Direct calculational methods in
homotopy theory were used by Barratt, Jones, and Mahowald to construct a
cell complex that can be used to solve the homotopy theory problem and prove
that such manifolds also exist in dimensions 30 and 62. I believe a construction
of such a manifold has been worked out in dimension 30, but that has certainly not
been done in dimension 62. Periodicity phenomena play a huge role in modern stable
homotopy theory, and a crucial feature of the Hill, Hopkins, Ravenel proof is a
periodicity of order $2^8 = 256$. That enables them to solve the stable homotopy problem
and prove there is no manifold of Kervaire invariant one for $j\geq 8$. The reasons $j=7$
is so hard are several. Nobody has a really good reason for guessing which way the answer
will go. There is no reason to expect a relevant periodicity of order $2^7$. Direct calculation
of the Adams spectral sequence through dimension $126$ is just plain hard: the
calculations blow up. There is a chance that the methodology of Barratt, Jones,
and Mahowald might extend to prove existence (if that is how the answer turns out!),
but it will probably be much harder to prove nonexistence (if that is the answer).

quid, that is a great resource, exactly what I was seeking. And thanks,Fernando, for the reference to (I guess?): Snaith, Victor P. (2009), "Stable homotopy around the Arf-Kervaire invariant,"Progress in Mathematics, 273, Birkhäuser Verlag. – Joseph O'Rourke Jul 1 '12 at 1:31