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It is known that only the complete graphs on 4 vertices or fewer are planar graphs, and all complete graphs $K_n$ where $n \ge 5$ are not planar.

It is also known only the general quartic, cubic, quadratic, and linear equations are always generally solvable with a finite number of algebraic steps. Polynomials with degree 5 or higher are not solvable in this manner (there is no general quintic formula). This is true for polynomials of degree n where $ n \ge 5$.

Is there is some relation between the nature of complete graphs on 5 or more vertices not being planar and the fact that polynomials of degree 5 or greater are not solvable? If so, what is this relation?

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    Well the automorphisms group of $K_5$ is $S_5$ – 2017-02-12

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I guess there is no relation. And the reason is following. Planaristy of graph is just a special case of genus $0$. Genus of $K_5$ is 1. I don't know anything similar about equations of degree $5$, they are just not solvable by radicals.