Distribution of the number of bits changed during generation of all binary tuples
It is somewhat more natural to normalize the number of bit changes per step by the number of steps, namely $2^n-1$. We can then rephrase your question as follows. Let $\pi$ be a random permutation of $\{0,1\}^n$. You are interested in the distribution of
$$
X = \frac{1}{2^n-1} \sum_{i=0}^{2^n-2} d(\pi(i),\pi(i+1)),
$$
where $d$ is Hamming distance. We can compute the moments of $X$ using linearity of expectation. The expected value of $X$ is
$$
\mathbb{E}[X] = \mathbb{E} d(\pi(0),\pi(1)) = \mathbb{E}_{x \neq 0} d(0^n,x) = \frac{\frac{n}{2} 2^n}{2^n-1} \approx \frac{n}{2}.
$$
The second moment of $X$ is
$$
\mathbb{E}[X^2] = \frac{1}{2^n-1} \mathbb{E} d(\pi(0),\pi(1))^2 + \frac{2(2^n-2)}{(2^n-1)^2} \mathbb{E} d(\pi(0),\pi(1)) d(\pi(1),\pi(2)) + \frac{(2^n-2)(2^n-3)}{(2^n-1)^2} \mathbb{E} d(\pi(0),\pi(1)) d(\pi(2),\pi(3)) \approx
\mathbb{E} d(\pi(0),\pi(1)) d(\pi(2),\pi(3)) \approx \left(\frac{n}{2}\right)^2.
$$
More generally, for every fixed $t$ we have
$$
\mathbb{E}[X^t] \approx \left(\frac{n}{2}\right)^t.
$$
This suggests that $X$ is tightly concentrated around $n/2$.

Going further, one is tempted to conjecture that the distribution is roughly normal, since the dependence between non-adjacent summands is very slight. This can be verified or refuted by repeating the calculation above with $d(\pi(i),\pi(i+1))$ replaced by $d'(\pi(i),\pi(i+1)) = d(\pi(i),\pi(i+1)) - n/2$. Since $d(\pi(i),\pi(i+1))$ is distributed almost like $\mathrm{Bin}(n,1/2)$, this suggests that $X$ is distributed almost like $N(n/2,n/4)$, a normal random variable with mean $n/2$ and variance $n/4$.