Solution:

The $32$ subsets split into $16$ complementary pairs $\{X, X^{\mathsf{c}}\},$ and no collection can contain both members of a pair (they are disjoint). A collection of $16$ pairwise-intersecting subsets must therefore contain exactly one member of every pair; in particular it contains $\{1,2,3,4,5\}$ and not $\emptyset.$

If some singleton $\{x\}$ is chosen, every member must meet $\{x\},$ i.e. contain $x.$ Exactly one set in each complementary pair contains $x,$ so the collection must be exactly the $16$ subsets containing $x:$ this gives $5$ collections. Otherwise no singleton is chosen, so all five $4$-element sets are in the collection. Any two $3$-element subsets of a $5$-element set intersect, a $4$-element set is disjoint only from its complement, and a chosen $2$-element set and a chosen $3$-element set are disjoint only if they are complements, which cannot both be chosen. So the only remaining condition is that the chosen $2$-element sets pairwise intersect.

Viewing $2$-element sets as edges of $K_5,$ a pairwise-intersecting collection of edges either has all edges through one common vertex or is a triangle. The number of such edge families is: the empty family ($1$), triangles ($\binom{5}{3} = 10$), and nonempty families within a star, $5(2^4 - 1) - 10 = 65$ (subtracting the $10$ single edges counted at both endpoints). That is $1 + 10 + 65 = 76$ collections, for a total of $5 + 76 = 81.$