If $S_n(a,b,c)$ is the number of solutions with $\sum_{i=1}^n f_i = a$, $\sum_{i=1}^n g_i = b$ and $\sum_{i=1}^n f_i g_i = c$, then we have $ S_n(a,b,c) = \sum_{j=0}^{\min(4,a)} \sum_{k=0}^{\min(4,b,\lfloor c/j \rfloor )} S_{n-1}(a-j,b-k,c-jk)$ (where we ignore the $\lfloor c/j \rfloor$ if $j=0$). The initial condition is $S_0(a,b,c) = 1$ for $a=b=c=0$, $0$ otherwise. On the other hand, the number of solutions with $\sum_{i=1}^n f_i = a$, $\sum_{i=1}^n g_i = b$ and no condition on $\sum_{i=1}^n f_i g_i$ is $A_n(a) A_n(b)$ where $A_n(a) = \sum_{j=0}^{\min(4,a)} A_{n-1}(a-j)$, and $A_0(a)=1$ for $a=0$, $0$ otherwise.
Your number is then $F_n = A_n(6) A_n(5) - \sum_{c=0}^4 S_n(6,5,c)$. I used the following Maple program:
A := proc (n, a) local j; option remember; if n = 0 then if a = 0 then 1 else 0 end if else add(A(n-1,a-j),j = 0 .. min(4,a)) end if end proc; S := proc (n, a, b, c) local j, k; option remember; if n = 0 then if a+b+c = 0 then 1 else 0 end if else add(S(n-1,a,b-k,c),k = 0 .. min(4,b)) +add(add(S(n-1,a-j,b-k,c-j*k),k = 0 .. min(4,b,floor(c/j))),j = 1 .. min(4,a)) end if end proc; F:= n -> A(n,6)*A(n,5) - add(S(n,6,5,c),c=0..4); L:= [seq(F(n),n=0..11)];
In practically no time, Maple produces $ L := [0, 0, 12, 318, 2780, 14515, 55056, 167664, 435456, 1003014, 2104140, 4096422]$
It can then find a polynomial to interpolate these:
CurveFitting:-PolynomialInterpolation([$0..11],L,n);
$ {\frac {19}{144}}\,{n}^{7}+{\frac {497}{240}}\,{n}^{6}-{\frac {2723}{ 144}}\,{n}^{5}+{\frac {3559}{48}}\,{n}^{4}-{\frac {1361}{9}}\,{n}^{3}+ {\frac {9107}{60}}\,{n}^{2}-58\,n $
which agrees, of course, with Yuval's result.