Sorry, I came across this 1 year old question only now ...
In fact, there are lots of results showing that explicit graphs with some properties imply strong lower bounds for boolean functions. Say, graphs of high affine or projective dimension imply strong lower bounds for formulas and branching programs. There are also "simpler" measures of graphs, good lower bounds on which would have great consequences in computational complexity. Let me sketch some of them.
View graphs as sets of edges. Let $s(G)$ be the smallest number $s$ such that $G$ can be written as an intersection of $\leq s$ graphs, each of which is a union of $\leq s$ bicliques (bipartite complete graphs). Easy counting shows that $s(G)\geq n^{1/2}$ for almost all bipartite $n\times n$ graphs. But by Valiant's results, every explicit bipartite graph $G$ (more exactly, a sequence of graphs) with $s(G)\geq n^c$ for a constant $c>0$ would resolve an old problem: would give a boolean function which cannot be computed by a log-depth circuit of linear size. It is conjectured that dense graphs without $K_{2,2}$ have large $s(G)$.
Even better, let $Star(G)$ be the smallest number of fanin-$2$ union and intersection operations that are enough to generate $G$ starting with complete stars (graphs of the type $K_{1,n}$ or $K_{n,1}$). Counting shows that most of the graphs have $Star(G)=\Omega(n^2/\log n)$. But any $G$ with $Star(G)\geq (4+c)n$ for a constant $c>0$ would give an explicit boolean function requiring circuits of exponential size! If the graph has dimension $m\times n$ with $m=o(n)$, then even a lower bound $Star(G)\geq (2+c)n$ would have the same consequences. The best we can show so far is $Star(G)\geq 2n-1$.
Let $Sym(G)$ be the smallest number $t$ for which there exists a subset $T\subseteq\{0,1,\ldots,t\}$ and a sequence of $t$ bicliques such that $(u,v)\in G$ iff the number of bicliques containing $(u,v)$ belongs to $T$. Again, counting gives $Sym(G)\geq n/2$ for most of the graphs. But by results of Yao, Beigel and Tarui any explicit graph with $Sym(G)$ larger than $2^{poly(\ln\ln n)}$ would give us a boolean function outside $ACC$. Warning: being "combinatorialy complicated" alone does not imply large $Sym(G)$: there exists strongly Ramsey graphs for which $Sym(G)=O(\log n)$, even if $T$ = set of odd integers.
More details on how all this happens can be found here.