I have been looking for possible counter-examples in $\left(\mathbb{Z}/n \mathbb{Z},+\right)$ for different values of $n$ and different subsets $A$ and $B$ using a SAGE program that I give below.
I have found one modulo 7 with
$A = [0, 1, 3]$ and $B = [0, 3, 4, 5]$.
Indeed :
$A+B = [0, 1, 3, 4, 5, 6]$
$A+A = [0, 1, 2, 3, 4, 6]$
$A-B = A-A = B+B = B-B = [0, 1, 2, 3, 4, 5, 6]$
giving a LHS $(6 \times 7)^2$ less that the RHS $6 \times 7 \times 7\times 7$ in your formula.
I have found another counterexample modulo 8 :
with $A = [0, 1, 2, 5]$ and $B = [0, 4, 5, 7]$.
... and one modulo $12$ :
with $A = [0, 1, 3, 6, 9, 10, 11]$ and $B = [1, 7, 8, 10, 11]$.
and also mod 14 with :
$A=[2,4,5,9,10,12]$ and $B=[0,3,5,6,7,9,12]$.
Besides, your inequality is verified in almost all cases with, sometimes, an equality between the LHS and RHS.
Here is the SAGE program :
from sage.misc.prandom import randrange
n=6
R=Integers(n); # integers mod n
Li=list(R)
X=Set([0,1,2,3,4,5]); # n elements
Y=X.subsets() # list of all the subsets of X
ne=Y.cardinality() # ne=2^n
def selec(m) : # getting the m-th subset
A=[];
for x in Y[m] :
A.append(Li[x])
return list(A)
def ol(L) : # opposite of a list
mL=[]
for k in Li :
mL.append(-k)
return(mL)
def cs(A,B) : cardinality of the sum of 2 subsets
C=[]
for a in A :
for b in B :
c=a+b
if not(c in C) :
C.append(c)
return len(C)
# Random selection of two subsets :
m=randrange(ne);A=selec(m);show("A = ",A)
m=randrange(ne);B=selec(m);show("B = ",B)
C=[cs(A,B),cs(A,ol(B)),cs(A,A),cs(A,ol(A)),cs(B,B),cs(B,ol(B))]
show("Cardinalities of A+B, A-B, etc. : ",C)
r=(C[0]*C[1])**2-C[2]*C[3]*C[4]*C[5];show("Diff, between LHS and RHS = ", r)