SNR improvement from placing analog filter ahead of digital filter
No, you can't double-dip like that.

While it is true that a lowpass integrating stage that reduces noise bandwidth by a factor of \$N\$ does improve linear SNR by \$\sqrt{N}\$, you cannot simply cascade two stages to get an SNR improvement of \$N\$.  Here's why:

The SNR improvement from averaging \$N\$ consecutive samples is conditional on the characteristics of sampled additive white Gaussian noise, specifically that they are independently distributed.  The following relationship holds for the sum of independent Gaussian random variables:

If

$$Y = \sum_i X_i$$

with

$$X_i \sim \mathcal{N}(\mu_i, \sigma_i^2) \quad \forall i$$

then

$$Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$$

where

$$\mu_y = \sum_i \mu_i$$

and

$$\sigma_y^2 = \sum_i \sigma_i^2$$

If the variables are i.i.d, which is to say that \$\mu_i = \mu_0 \; \forall i\$ and \$\sigma_i = \sigma_x \; \forall i\$, then averaging gives

$$\frac{1}{N} \sum_{i=1}^N X_i \sim \mathcal{N}\left(\mu_x,\left(\frac{\sigma_x}{\sqrt{N}}\right)^2\right)$$

As the acquisition system DC offset is properly zeroed, \$\mu_x = 0\$.  In this case the noise is reduced by a factor of \$\sqrt{N}\$ by averaging.  The same result holds for averaging in the analog domain via integration.

However, this result only applies when the Gaussian random variables are independent (white noise).  As soon as a lowpass filter is added in front of the averaging block, the noise contributions become highly correlated, and integration of a signal which is already band-limited results in no additional reduction.