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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).