After trying various formulas involving combinations, and mulltiplications by the n-fold symmetry and trying to subtract out the rectangles, which would otherwise be counted twice, and differentiating odd n from even n, I gave up trying to find a formula.

The program keeps track of the slope of every line, identifying each line by the ID of its two defining points (ID being the subscript on the table of point coordinates).

That table of slopes is sorted to find all pairs of lines with matching slope; each such map goes onto a table of quadrilaterals, which are all trapezoids. Duplicates are discarded and unique sets of four points remain. This set is counted and reported.

clearvars,clc

for n=4:15

  pts=double.empty(0,2); % x,y

  incr=360/n;

  for a=.1:incr:360-incr+.2

    pts(end+1,:)=[cosd(a),sind(a)];

  end

  lines=double.empty(0,3); % v1,v2,slope

  for v1=1:n

    for v2=v1+1:n       

        slope=(pts(v2,2)-pts(v1,2))/(pts(v2,1)-pts(v1,1));

        lines(end+1,:)=[v1,v2,slope];      

    end

  end

  lines=sortrows(lines,3);

  quad=double.empty(0,4);

  for l1=1:size(lines,1)

    for l2=l1+1:size(lines,1)

      if abs(lines(l1,3)-lines(l2,3))<abs((1e-10)*lines(l2,3))

        quad(end+1,:)=sort([lines(l1,1) lines(l1,2) lines(l2,1) lines(l2,2)]);

      end

    end

  end

  quad=unique(quad,'rows');

  fprintf('%3d %4d\n',n,size(quad,1))

end

finding the numbers:

  n  trapezoids from
      n-gon vertices
  4    1
  5    5
  6    9
  7   21
  8   30
  9   54
 10   70
 11  110
 12  135
 13  195
 14  231
 15  315