Suppose that there exists a $n-$dimensional vector $(1, 1,\dots, 1)$ which are all $1$. For this vector doing the following: at each step, choose one of $i$-th coordinate uniformly at random and flipped, i.e. turns $1$ into $0$ and vice-versa. Clearly, this is a Markov chain.
What is the long-run proportion of times the vector is still $(1, 1,\dots, 1)$?
It is easy to show that this MC is irreducible. I try to assume that the $X_n$ is the number of $1$ in the vector after $n$ flips. So how to get $$\pi_n =\lim_{k\to\infty}\frac{1}{n}\sum_{m=1}^kP_{nn}^m?$$ where $P_{nn}^m=\mathbb{P}(X_m=n\mid X_0=n)$.
Is there any more direct method to get the result $1/2^n$? I feel like this chain similar to the expectation of a simple randon walk return to its starting point.
Observe that for any given sequence of flips, we land back on our original vector if and only if each coordinate is flipped an even number of times (possibly $0$). It follows that after an odd number of flips, we can never land back on the original vector. We will hence consider only sequences where $L=2k$ is even.
For any length $L=2k$, the total number of sequences of flips is $n^{2k}$. Let's count the number of sequences of flips which satisfy the parity requirement.
Let $2v_i$ be the number of flips at position $i$. We must have
$$\sum_{i=1}^nv_i = k \tag{$1$}$$
where $0\leq v_i \leq k$ are integers. Now, for each solution to $(1)$, we can order the flips in
$$\binom{2k}{2v_1,2v_2,\dots,2v_n}$$
distinct ways. This would give the total number of sequences as
$$\sum_{v_1+v_2+\dots+v_n = k} \binom{2k}{2v_1,2v_2,\dots,2v_n} \tag{2}.$$
This answer shows that $(2)$ can be rewritten as
$$\frac1{2^n }\sum_{m=0}^n \binom{n}m{(n-2m)}^{2k}\tag{3},$$
and hence the probability that after a sequence of $2k$ flips we are back to the original vector is
$$p_{2k} = \frac{\sum_{m=0}^n \binom{n}m{(n-2m)}^{2k}}{2^n n^{2k}}.$$
The quantity you seek can be explicitly written as $\lim_{k\to\infty} \frac1{2k} \sum_{j=1}^k p_{2j}$. I haven't really found a way to show that it reduces to $1/2^n$ yet.
EDIT: Rewrite
$$p_{2k} = \frac1{2^n}\sum_{m=0}^n \binom{n}m{\left(1-\frac{2m}n\right)}^{2k},$$
and reorder the summations in the limit expression to obtain
$$\frac1{2k\cdot2^n}\left(\sum_{m=0}^n\binom{n}m \sum_{j=1}^k {\left(1-\frac{2m}n\right)}^{2j}\right). \tag{4}$$
Separately sum the indices $m=0$ and $m=n$ to obtain
$$\frac1{2k\cdot2^n}\left(2k+\sum_{m=1}^{n-1}\binom{n}m \sum_{j=1}^k {\left(1-\frac{2m}n\right)}^{2j}\right) \\= \frac1{2^n} + \underbrace{\frac1{2k\cdot2^n}\left(\sum_{m=1}^{n-1}\binom{n}m \sum_{j=1}^k {\left(1-\frac{2m}n\right)}^{2j}\right)}_{\alpha}.$$
We show that as $k\to\infty$, $\alpha$ approaches $0$.
Notice that for $1\leqslant m \leqslant n-1$, we have ${\left(1-\frac{2m}n\right)}^{2j} \leqslant {\left(1-\frac2n\right)}^{2j}$. Let $t = {\left(1-\frac2n\right)}^2$ and notice that $|t|<1$ and that $t$ depends only on $n$. Write
\begin{align} 0\leqslant \alpha &\leqslant \frac1{2k\cdot 2^n}\sum_{m=1}^{n-1}\binom{n}m \sum_{j=1}^k t^j \\&= \frac1{2k\cdot 2^n}\sum_{m=1}^{n-1}\binom{n}m \frac{t(1-t^k)}{1-t} \\&= \frac1{2k\cdot 2^n}\cdot (2^n-2) \cdot \frac{t(1-t^k)}{1-t} \\&\leqslant \frac1{2k} \cdot \underbrace{\frac{t(1-t^k)}{1-t}}_{\beta}. \end{align}
As $k\to\infty$, $\beta$ approaches $t/(1-t)$ and in particular remains bounded. It follows from the squeeze theorem that $\alpha$ indeed approaches $0$ as $k\to\infty$, and hence $\pi_n = 2^{-n}$.