How annoying is my Euler diagram?
MATLAB - 109 Bytes
Try it online
function n=E(s)
i=2;k=1;n=0;while i&lt;numel(s)
k=mod(k,4)+1;n=n+s(i)*((k==1)-(k==3)/2);i=i+s(i)+1+(k&lt;2);end

ungolfed:
function n=E(s)             % function def, input 2 and output n
i=2;                        % sequence index
k=1;                        % 1=sub,2=equivalent,3=super,4=part
n=0;                        % initialized output
while i&lt;numel(s)            % Loop through sequence
k=mod(k,4)+1;               % if k==4, then a new set has started, so back to 1.
n=n+s(i)*((k==1)-(k==3)/2); % s(i) represents n,m,p,q  
                            % If k==1, then s(i)=q, so add s(i) to n. 
                            % If k==3, s(i)=m, so subtract s(i)/2 from n.  
i=i+s(i)+1+(k&lt;2);           % Advance index by (n,m,p,q) and 1 if k==1 (aka k&lt;2) to skip set number definition. 
end                         % end

Explanation:
This may be a naive approach, but from what I see, the only way to have a perimter intersection is when you have a set as a part of another.
For super/subsets, there will be no perimeter intersections, since they will be concentric.
For sets that match exactly, there will only be one shape that represents two sets, so there are no perimeter crossings between them.
Therefore, the only time you can have an overlap is for parts. From there it is simple: every crossing includes two points, since by the nature of overlapping rectangles, there will be two and only two intersections of the perimeters. And since the sets that contain parts of another must reference each other (i.e. if set 1 and 2 overlap, set 1 will have 2 in its part input, and 2 will have 1), there will be 2*(number of sets that are part of another). However, sets that are exactly equal only produce one shape, so any crossings between this shape and another must be reduced according to the number of sets that are equivalent. Thus, the sum of the number of elements that fall into the &quot;parts&quot; category across all sets equals the number intersections.
Mathematically,
For \$k\$  sets
\$ n_{crossings} = \displaystyle \sum_{i=1}^k(q_i - m_i/2) \$