If G is formed from a smaller graph H that is not a clique by adding two vertices x and y, such that x and y are not adjacent to each other but adjacent to all other vertices of G, then $tw(G)=tw(H)+2$. For, in any tree decomposition of $G$, either $x$ and $y$ have disjoint subtrees or they have overlapping subtrees. If they have disjoint subtrees, all the other subtrees have to include the shortest path between the trees for $x$ and $y$, from which it follows that the treewidth is $n-2$; the assumption that $H$ is not a clique can then be used to show that $n-2\ge tw(H)+2$. Alternatively if $x$ and $y$ have overlapping subtrees, every other vertex has to have a subtree that touches the intersection of the two subtrees of $x$ and $y$, and we can restrict the tree decomposition to that intersection, giving a tree decomposition in which $x$ and $y$ participate in every tree node.
This implies that the hyperoctahedral graph $K_{2,2,2,\dots}$ with $2k$ nodes is a minimal forbidden minor for width $2k-3$. For, the octahedral graph $K_{2,2,2}$ is a minimal forbidden minor for width three, from which the argument above shows that the hyperoctahedral graph has width $2k-2$. And if any edge contraction or edge deletion is performed in the hyperoctahedral graph, the symmetries of the graph allow us to assume that the operation is happening to one of the twelve edges in the base octahedron, causing its width and the width of all the hyperoctahedra built from it to decrease.
(The other class of graphs you should be including in your question along with the complete graphs are the grid graphs. An $r\times r$ grid has treewidth $r$. It's separate from the complete graph minors because its planar and therefore has no complete minor with more than four vertices. It's not a minimal forbidden minor, though, because some small changes (such as contracting the corner vertices) don't change its treewidth.)
