The answer is $\boxed{2^{22}}$. We may assume that each dwarf has a set of hats with different colors. Consider a bipartite graph $G=(C\cup D, E)$ where $C,D,E$ represents the set of colors, dwarfs and hats respectively. Draw an edge between $c\in C$ and $d\in D$ iff the dwarf $d$ has a hat in color $c$, then we have $$\#C=\#D=66,\; \#E=111.$$Note that each possible festivity corresponds to a perfect match in $G$, therefore our goal is to estimate the maximum possible number of perfect matches in $G$. Lemma. For any bipartite graph $G=(C\cup D, E)$ where $\#C=\#D=n,\#E=e$, $G$ contains at most $2^{\left\lfloor\frac{e-n}{2}\right\rfloor}$ distinct perfect matches. Proof. Apply induction on $n$. For $n=1$ the conclusion is trivial. Suppose the conclusion holds for $n-1$, consider the case of $n$. WLOG assume that each vertex has a positive degree, then we can pick out the vertex $v$ with the smallest degree. Let the neighbors of $v$ be $u_1,\dots,u_s(s\ge 1)$. For each possible perfect match $M$, let the neighbor of $v$ in $M$ be $u_i$. Wipe out $v,u_i$ (and all their adjacent edges, but do not wipe out their other neighbors) from $G$ we are left a graph $G'$ with $2(n-1)$ vertices and at most $e-(2s-1)$ edges (due to the minimality of $s$). From the induction hypothesis there are at most $$2^{\left\lfloor(e-2s+1-n+1)/2\right\rfloor}=2^{-s+1}\cdot2^{\left\lfloor(e-n)/2\right\rfloor}$$perfect matches in $G'$, each of which corresponds to a $M$ with $(v,u_i)\in M$. Since there are $s$ possible choices for $i$, the total number of perfect matches does not exceed $$s\cdot 2^{-s+1}\cdot2^{\left\lfloor(e-n)/2\right\rfloor}\le2^{\left\lfloor(e-n)/2\right\rfloor}.$$Take $n=66, e=111$ we get that at most $2^{22}$ festivities can be organized. Construction for $\bold{2^{22}}$ festivities. Let $$C=\{c_1,c_2,\dots,c_{66}\}, D=\{d_1,d_2,\dots,d_{66}\},$$then the $111$ edges are chosen as follows. $$(c_i, d_i)\;(1\le i\le 66);$$$$(c_{2i-1},d_{2i}),(c_{2i},d_{2i-1})\;(1\le i\le 21);$$$$(c_{43}, d_{44}),(c_{44}, d_{45}),(c_{45}, d_{43}).$$It's easy to check that there are exactly $111$ edges and $2^{22}$ perfect matches.

We will solve the problem with generating functions. The answer is $2^{22}$ Let $c_j$ denote a hat of color $j$. Replace $66$ with $n$ and $111$ with $k$. I will prove if we have $n$ linear polynomials ($P_j$ in $c_1, \cdots, c_n$ for $j=1,2,\cdots,n$) with only 0,1 as coefficients and a total of $k$ terms then $[\prod\limits_{j=1}^n c_j] \prod\limits_{j=1}^n P_j(c_1,\cdots,c_n) \le 2^{\lfloor \frac{k-n}{2} \rfloor}$. We proceed by induction, with the base cases $n=1,2$ verifiable by hand. Inductive step: We have the following recursive structure: if a term $c_t$ appears in $m$ polynomials, then if it is the only term in a polynomial then it must be used, and if only one polynomial has it it must also be used. Therefore, $m\ge 2$ and this implies $f(t,k) \le mf(t-1,k-m-1)$. If $m=2$ we have $f(t,k)\le 2f(t-1,k-3)$ so we are done by inductive hypothesis. Same for $m=4$ and any $m\ge 6$. We are done by inductive hypothesis unless $m=3$ or $m=5$ for all coefficients $c_t$. Therefore, each term is in 3 or 5 polynomials. This means $k-t$ is odd. Now, if say $c_t$ shows up in 3 polynomials, call $P,Q,R$. Then if two of P,Q,R has at least $3$ terms, then picking $c_t$ from $P$ with at least 3 terms suggest there are at most $2^{\lfloor \frac{k-t-4}{2} \rfloor} = \frac 14 2^{\lfloor \frac{k-t}{2} \rfloor}$ ways to picking $[c_1c_2\cdots c_t]$ coefficients. This implies that if $c_t$ shows up in 3 or 5 polynomials, then at most one of them can have more than 2 terms. Take a polynomial containing two terms, call $c_j, c_k$. Since $c_j$ appears in at least 3 polynomials, it follows that if I pick $c_j$ there exists at most $2^{\lfloor \frac{k-t-3}{2} \rfloor}$ ways because I get rid of $c_j, c_k$ and the other two occurences of $c_j$. The same conclusion holds for $c_k$, completing the induction.

Construction for $n=66, k=111$. We take 22 monomials $c_j$ where $j=45,\cdots, 66$, and 2 copies of $c_{2i-1}+c_{2i}$ where $i=1,\cdots,22$. For each $i=1,\cdots,22$ there are 2 ways to pick $c_{2i-1}c_{2i}$, for a total of $2^{22}$ ways. Since $22=\lfloor \frac{111-66}{2} \rfloor$, the conclusion follows.