Here's an almost complete answer to the claim: No convex 7+ sided polygon can tessellate the plane. We don't need to compute interior angles or averages. We only need to count the number of interior angles in two ways.
We'll use contradiction of course. Let our polygon $P$ have $n$ sides and let it have area 1. Take a circle of area $A$. Keep every copy of $P$ that lies at least partially in the circle.
I will assume, without careful proof, that the number of polygons in our circle, which can be considered faces in the context of the Euler characteristic of the resulting planar graph, is $F=A+O(A^{1/2})$.
Now I will also assume, without careful proof, that the number of edges in our resultant graph is $E=nF/2+O(A^{1/2})$, as every edge is shared by two polygons except for the $O(A^{1/2})$ on the boundary. Thus $E=nA/2 + O(A^{1/2})$.
Using the Euler characteristic equation for a planar graph, we can compute $V = E-F+1 = \left(\frac{n}{2}-1\right)A + O(A^{1/2}).$
Now we can count the number of interior angles in two ways. First, it is easily seen that the number must be $nF$. But also,
$nF = \sum_{\mbox{v in all vertices}} \mbox{(# of angles meeting at v)} \\ = \sum_{\mbox{v in interior vertices}} \mbox{(# of angles meeting at v)} + \sum_{\mbox{v in edge vertices}} \mbox{(# of angles meeting at v)}.$
The number of edge vertices $V_e$ is assumed to be $O(A^{1/2})$ so that the number of interior vertices $V_i$ is assumed to be the same as $V=(n/2-1)A+O(A^{1/2})$
Because the number of angles meeting at a vertex must be bounded above given any polygon $P$, the second term in our split sum must be $O(A^{1/2})$.
Now using convexity, we note that for interior angles, the number of interior angles meeting at any interior vertex must be greater than or equal to 3. So,
$nF \geq 3V_i + O(A^{1/2}) = 3V + O(A^{1/2}).$
Substituting gives
$nA \geq 3(n/2-1) A + O(A^{1/2}),$
which cannot hold in the limit as $A \to \infty$ when $n > 6$. This gives us our desired contradiction.