I am sorry to say that the answer to Question 1 is NO.
As you state, $q=(3-n)p/(2n)$ gives a solution to the problem. Thus the quartic must be birationally equivalent to an elliptic curve.
Using standard methods, it is possible to show that this elliptic curve is of the form \begin{equation*} G^2=H^3+25( h_1 H + h_2)^2 \end{equation*} where \begin{equation*} h_1=(n^2+3) \hspace{1cm} h_2=4(n^2+1)(n^2+2n+5)(n^2-2n+5) \end{equation*} with the reverse transformation \begin{equation*} \frac{p}{q}=\frac{2n(G+(n^2+11)H+20(n^6+7n^4+31n^2+25)}{(3-n)(G+(3n^2+4n+5)H+20(n^6+7n^4+31n^2+25))} \end{equation*}
The elliptic curves have torsion points when $H=0$, and the torsion subgroup seems to be $\mathbb{Z}3$.
Applying the Birch and swinnerton-Dyer conjecture to these curves, the first positive rank is at $n=12$ as you state, but the second is at $n=13$, where $H=-608000/9$ gives a rational point. Substituting backwards we arrive at the solution to the initial problem for $n=13$: $a=-1181$, $b=-6033$, $c=6302$, $d=6805$, $e=-4702$ and $f=-3015$.
I computed height estimates for $n$ up to $100$. There seems no obvious pattern and some of the heights are very large, suggesting a solution but one with enormous numbers. Since the elliptic curves do not have a point of order $2$, finding these solutions will not be easy.