Consider the following matrices
$$I = \begin{pmatrix}1& 0\\ 0 & 1\end{pmatrix},\quad X = \begin{pmatrix}0& 1\\ 1 & 0\end{pmatrix}$$
Define the indexed quantity $P^n_k(I, X)$ to be the sum of all possible $n$-fold tensor products of $I$ and $X$ with $k$ factors of $X$. For instance, $P^3_1(I,X) = I\otimes I\otimes X + I\otimes X\otimes I + X\otimes I\otimes I$.
Let the $2^n\times 2^n$ matrix with all entries equal to $1$ be denoted by $U_n$. Then
$$U_n = \sum_{k=0}^n P^n_k(I,X).$$
I have verified that this is numerically true. For $n=1$, we have
$$U_1 = \begin{pmatrix}1& 1\\ 1 & 1\end{pmatrix} = I + X.$$
For $n=2$
$$U_2 = \begin{pmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{pmatrix} = I\otimes I + I\otimes X + X\otimes I + X\otimes X$$
and so on. How can one prove this result for general $n$?
Using your notation, we have a formula as follows: $$I\otimes P_{k}^{n}+X\otimes P_{k-1}^{n}=P_{k}^{n+1},k\ge 1$$ where I use $P_{k}^{n}$ for shorthand of $P_{k}^{n}\left( I,X \right) $.
And mind that $U_{n+1}=I\otimes U_{n}$, so we have $$\begin{align*} U_{n+1}&=(I+X)\otimes \sum_{k=0}^n{P_{k}^{n}} \\ &=I\otimes \sum_{k=0}^n{P_{k}^{n}}+X\otimes \sum_{k=0}^n{P_{k}^{n}} \\ &=P_{0}^{n+1}+I\otimes \sum_{k=1}^n{P_{k}^{n}}+X\otimes \sum_{k=0}^{n-1}{P_{k}^{n}}+P_{n+1}^{n+1} \\ &=P_{0}^{n+1}+\sum_{k=1}^n{P_{k}^{n+1}}+P_{n+1}^{n+1} \\ &=\sum_{k=0}^{n+1}{P_{k}^{n+1}}\end{align*} $$ You already gave the case of $U_1$ and $U_2$ which satisfy what you want, and by induction we have the conclusion that $U_n$ also satisfy $U_n=\sum_{k=0}^{n}{P_{k}^{n}}$.