Bode magnitude computation
The amplitude of the bode-plot is found by traversing the imaginary axis, or

$$s = j\omega$$

So the formula for the amplitude should actually be

$$A_{dB} = 20\cdot \log_{10} \left(\left| \frac{10^6\cdot j\omega}{j\omega + 1} \right| \right)$$

For \$\omega = 0.1\$ and \$\omega=1\$:

$$A_{dB}(0.1) = 20\cdot \log_{10}\left( \frac{10^6\cdot 0.1}{\sqrt{0.1^2 + 1^2}} \right) = 99.96\ dB$$

$$A_{dB}(1) = 20\cdot \log_{10}\left( \frac{10^6\cdot 1}{\sqrt{1^2+1^2}} \right) = 116.99\ dB$$

The 20dB/decade or 6dB/octave "rule" only holds asymptotically. The closer you get to the corner frequency, the worse the approximation is. In your case, that corner frequency is at \$\omega_c = 1\$, so the approximation will be at its worst.

The corner frequency is also the point where these asymptotes intersect. This is the point where the imaginary part of the denominator (\$j\omega + 1\$) starts to dominate over the real part, or \$|j\omega| = |1|\$, which is the case for \$A_{dB}(1)\$. You end up dividing by \$\sqrt{2}\$, which is approximately 3dB.