Solution:
Answer: $117649$

We first note that $x^{7}-7=\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{7}\right)$, which implies, replacing $x$ by $-x$ and taking the negative of the equation, that $\left(x+r_{1}\right)\left(x+r_{2}\right) \cdots\left(x+r_{7}\right)=x^{7}+7$. Also note that the product of the $r_{i}$ is just the constant term, so $r_{1} r_{2} \cdots r_{7}=7$.

Now, we have that
$$\begin{aligned} 2^{7} \cdot 7 \cdot K^{2} & =\left(\prod_{i=1}^{7} 2 r_{i}\right) K^{2} \\ & =\prod_{i=1}^{7} 2 r_{i} \prod_{1 \leq i<j \leq 7}\left(r_{i}+r_{j}\right)^{2} \\ & =\prod_{1 \leq i=j \leq 7}\left(r_{i}+r_{j}\right) \prod_{1 \leq i<j \leq 7}\left(r_{i}+r_{j}\right) \prod_{1 \leq j<i \leq 7}\left(r_{i}+r_{j}\right) \\ & =\prod_{1 \leq i, j \leq 7}\left(r_{i}+r_{j}\right) \\ & =\prod_{i=1}^{7} \prod_{j=1}^{7}\left(r_{i}+r_{j}\right) . \end{aligned}$$
However, note that for any fixed $i$, $\prod_{j=1}^{7}\left(r_{i}+r_{j}\right)$ is just the result of substituting $x=r_{i}$ into $\left(x+r_{1}\right)(x+r_{2}) \cdots(x+r_{7})$. Hence,
$$\prod_{j=1}^{7}\left(r_{i}+r_{j}\right)=r_{i}^{7}+7=\left(r_{i}^{7}-7\right)+14=14$$
Therefore, taking the product over all $i$ gives $14^{7}$, which yields $K^{2}=7^{6}=117649$.