Pods $A,$ $B,$ $C,$ and $F$ are all pairwise connected. Four numbers from $1$ through $7$ that are all at least $2$ apart must be $\{1,3,5,7\}$. Thus $D,$ $E,$ and $G$ get the even grades $\{2,4,6\}$.

Pod $G$ is connected to $A$ and $F$. If $G=4$, then $A$ and $F$ would have to be $1$ and $7$, leaving $B$ and $C$ as $3$ and $5$. But then $E$, which is connected to $C$ and $F$, cannot be either remaining even grade without being only $1$ away from one of them. So $G$ cannot be $4$.

If $G=2$, then $A$ and $F$ must be $5$ and $7$. If $F=5$, then $E$ cannot be $4$ or $6$, so $F=7$. Also, $C$ cannot be $3$, since then $E$ again cannot be $4$ or $6$. Therefore $C=1$ and $E=4$, giving $C+E+F=1+4+7=12$.

The case $G=6$ is symmetric, giving $C=7,$ $E=4,$ and $F=1$. The same sum is $12$.

Thus, A is the correct answer.