Calculating the true values of biquad filter coefficients from normalized form
Note that a biquad has $5$ degrees of freedom (not $6$), because $a_0$ can always be chosen as $a_0=1$ without loss of generality:

$$\begin{align}H(z)&amp;=\frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}\\&amp;=b_0\cdot\frac{1+\hat{b}_1z^{-1}+\hat{b}_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}},\qquad \hat{b}_1=\frac{b_1}{b_0},\;\;\hat{b}_2=\frac{b_2}{b_0}\tag{1}\end{align}$$

Eq. $(1)$ shows that $b_0$ is just an overall gain (for fixed values of $\hat{b}_1$ and $\hat{b}_2$), as you've noted yourself.

If you don't know the formula that produced the filter coefficients, there is generally no way to know the intended gain. As mentioned in Dan's answer, the gain is often chosen such that quantization effects are minimized. It can often be compensated for after all computations have been performed.

In some cases, however, it is possible to estimate the intended gain. E.g., if the biquad is a low pass filter, it is reasonable to assume that the original frequency response has a value of $1$ at DC. This means that

$$b_0\frac{1+\frac{b_1}{b_0}+\frac{b_2}{b_0}}{1+a_1+a_2}=1\tag{2}$$

Similarly, for a high pass filter a common scaling makes sure that the frequency response equals $1$ at Nyquist, which implies

$$b_0\frac{1-\frac{b_1}{b_0}+\frac{b_2}{b_0}}{1-a_1+a_2}=1\tag{3}$$

You can come up with similar guesses for the intended scaling for other standard filter types, like band pass filters (unity gain at the center frequency) and band stop filters (unity gain at DC and/or Nyquist).