Number of possible access structure with $n$ person
An access structure is just a function $f:\{0,1\} ^n\to \{0,1\}$. It takes a subset of $\{1,\ldots,n\}$ as input (represented as it's characteristic vector) and returns 1 if authorized and 0 if unauthorized. Additionally, the function must be monotone, meaning that $A\subseteq B \implies f(A) \le f(B)$.

The number of monotone Boolean functions over $n$ variables is given by the Dedekind numbers (OEIS). It is a nontrivial formula!

Edit: You are interested in $f$ which are sensitive to all $n$ inputs. Let's call $f$ insensitive to input $i$ if there is a function $g$ (defined on subsets of $\{1,\ldots, i-1,i+1,\ldots,n\}$) such that $f(A) = g(A \setminus \{i\})$ for all $A$. If this is not the case then we say that $f$ is sensitive to $i$. Now say that $f$ is $k$-sensitive if it is sensitive to $k$ of its inputs. You can count the number of $n$-sensitive monotone functions using the inclusion-exclusion principle.

The idea is that $M(n)$ ($n$th Dedekind number) includes things you don't care about like $(n-1)$-sensitive functions. There are ${n \choose 1} M(n-1)$ Monotone functions of $n$ variables that are (at most) $(n-1)$-sensitive, so subtract them off. But since $M(n-1)$ also "overcounts" all the $(n-2)$-sensitive functions, you've subtracted them off too much and need to add them back. There are ${n \choose 2} M(n-2)$ of these. Etc etc etc. You finally get the thing you want as $\sum_{i=0}^n (-1)^i {n \choose i}M(n-i)$.