Trouble with understanding moving average filter as an oversampling filter
A moving average over 6 samples will not add 6 bits of precision to the ENOB (equivalent number of bits) as stated in the question. In general, a moving average over $N$ samples will increase resolution by $log2(N)/2$, and for 6 samples this would be approximately 1.3 bits.

This approximation holds to the extent that the quantization noise on each sample is represented by a uniformly distributed white noise process (which holds well when there are a sufficient number of quantization levels and the sampling clock is not correlated to the signal being sampled). From statistics, when you add N independent random variables, the standard deviation increases at $\sqrt{N}$. Thus in the average where we also divide that sum by $N$ the standard deviation is scaled by $\sqrt{N}/N = 1/\sqrt{N}$.

This results in a 3 dB reduction in quantization noise for every doubling of samples in the average (from $20Log_{10}(1/N)$ since dB is $20Log_{10}(\sigma)$ where $\sigma$ is the standard deviation. As already detailed in the post linked below, each bit of precision represents a 6 dB reduction in quantization noise: 

What are advantages of having higher sampling rate of a signal?

A moving average over 4 samples would add 1 bit of precision, 16 samples would add 2 bits, etc.

To support the increased number of bits, the result from the addition of the samples must have that many number of bits which is also apparent to eliminate overflow conditions. Filtering a digital signal (which a moving average is) grows the signal! This is why we must use extended precision accumulators when filtering. We can scale the accumulated signal back after summing (for the average) while scaling the signal first prior to the summation, or equivalently scaling the coefficients, would result in further noise contribution.