1. We need the function $f_a(x)=\tan^{-1}(2x)-3ax+7$ to be non-decreasing on $\left(-\frac{\pi}{6},\frac{\pi}{6}\right).$

For a differentiable function, non-decreasing on an interval means $f_a'(x)\ge 0 \quad \text{for all } x \text{ in the interval.}$

2. Differentiate: $f_a'(x)=\frac{2}{1+4x^2}-3a.$

So we need $\frac{2}{1+4x^2}-3a\ge 0 \quad \forall x\in\left(-\frac{\pi}{6},\frac{\pi}{6}\right).$ That is, $3a\le \frac{2}{1+4x^2} \quad \forall x.$

Hence the maximum possible value of $a$ is obtained from the minimum of $\frac{2}{1+4x^2}$ on the interval.

3. Now, $\frac{2}{1+4x^2}$ is a decreasing function of $x^2$, so its minimum on $\left(-\frac{\pi}{6},\frac{\pi}{6}\right)$ occurs at the endpoints $x=\pm \frac{\pi}{6}$.

Thus,

Simplify: $\frac{2}{1+\frac{\pi^2}{9}}=\frac{2}{\frac{9+\pi^2}{9}}=\frac{18}{9+\pi^2}.$

Therefore, $3\bar a=\frac{18}{9+\pi^2}$ so $\bar a=\frac{6}{9+\pi^2}.$

4. Now compute $f_{\bar a}\left(\frac{\pi}{8}\right)=\tan^{-1}\left(2\cdot \frac{\pi}{8}\right)-3\bar a\cdot \frac{\pi}{8}+7.$

This gives $f_{\bar a}\left(\frac{\pi}{8}\right)=\tan^{-1}\left(\frac{\pi}{4}\right)-\frac{3\pi}{8}\cdot \frac{6}{9+\pi^2}+7.$ So, $f_{\bar a}\left(\frac{\pi}{8}\right)=\tan^{-1}\left(\frac{\pi}{4}\right)-\frac{9\pi}{4(9+\pi^2)}+7.$

5. Now compare with the options. Option A is $8-\frac{9\pi}{4(9+\pi^2)}.$ This would require $\tan^{-1}\left(\frac{\pi}{4}\right)+7=8,$ which is false.

Also, note that $\tan^{-1}\left(\frac{\pi}{4}\right)\ne 1$ and certainly not something that simplifies to make any listed option exact.

So the exact derived value is $\boxed{7+\tan^{-1}\left(\frac{\pi}{4}\right)-\frac{9\pi}{4(9+\pi^2)}}.$

6. Therefore, none of the given options matches the mathematically derived answer. The stored correct answer A appears inconsistent with the question as written.