Quantization SNR of bipolar signal
Your question appears to be about the signal to quantization noise ratio for uniform quantization. You have to count each bit, but that generally doesn't mean that the SNR is $6N$ dB, where $N$ is the number of bits. You normally have a constant offset:

$$\text{SNR}_{\text{dB}}\approx c+6N\tag{1}$$

where the constant $c$ depends on the Crest factor of the signal.

To give you an example, assume that the signal is in the range $[-x_p,x_p]$, and assume uniform quantization with $N$ bits. Then the quantization step size is

$$q=\frac{2x_p}{2^N}\tag{2}$$

If we further assume that the quantization error is uniformly distributed in the interval $[-q/2,q/2]$, the quantization noise power is

$$P_n=\frac{1}{q}\int_{-q/2}^{q/2}\xi^2d\xi=\frac{q^2}{12}=\frac13\frac{x_p^2}{2^{2N}}\tag{3}$$

With $x_{\text{RMS}}$ the RMS value of the signal, the signal to quantization noise ratio is given by

$$\text{SNR}=\frac{x^2_{\text{RMS}}}{P_n}=3\frac{x^2_{\text{RMS}}}{x^2_p}\cdot2^{2N}=\frac{3}{C^2_x}\cdot2^{2N}\tag{4}$$

where $C_x$ is the signal's Crest factor. From $(4)$, the SNR in dB is

$$\text{SNR}_{\text{dB}}=10\log_{10}\left(\frac{3}{C^2_x}\right)+20N\log_{10}2=10\log_{10}\left(\frac{3}{C^2_x}\right)+6.02\cdot N\tag{5}$$

which has the form of Eq. $(1)$.