I think this quickly leads to difficult questions.

Thinking of $B_n$ as the semidirect product $H \rtimes S_n$ where $H \cong C_2 \times \cdots \times C_2$, we can factorize each element of $B_n$ as $h t$ where $h \in H$ and $t \in S_n$. If $t$ has order $r$ then $(ht)^r = h'$ for some $h' \in H$, of order at most $2$. Choosing $h$ suitably, we can assume $h'$ is not the identity. Hence the maximum order of an element of $B_n$ is twice the maximum order in $S_n$, say $m_n$. Landau proved that

$$\log m_n \sim \sqrt{n} \log n \quad\text{as $n \rightarrow \infty$.}$$

For $k$ reasonably large (but less than $n$), my guess is that this means there is an element of $B_n$ having a cycle of length $2m_n$ in the action of $B_n$ on $k$-faces.

If we position the hypercube so that its $(n-1)$-faces are determined by the equations $x_i = 1$ or $x_i = -1$ for $i \in \{1,\ldots, n\}$ then the $k$-faces correspond to equations $x_{i_1} = \pm 1, \ldots, x_{i_{n-k}} = \pm 1$, with $n-k$ chosen signs. Hence the stabiliser of a $k$-face is a conjugate of $B_k \times S_{n-k}$. Therefore, restricting the action of $B_n$ on $k$-faces to the chosen top group $S_n$, we get $S_n$ acting on the cosets of $S_k \times S_{n-k}$, i.e. $S_n$ acting on $k$-subsets of $\{1,\ldots, n\}$. For this action, I would guess that for 'most' $k$, the maximum cycle length is $m_n$. As far as I know, no result of this form has been proved.

Edit Let $c_n$ be the number of distinct prime factors of $m_n$. In Proposition 7 in Ordre maximal d'un élément du groupe symmétrique, Bull. Soc. Math. France. 97 (1969) 129–191, Nicolas proved that

$$ c_n \sim 2 \sqrt{\frac{n}{\log n}}.$$

Note that $c_n$ is the number of disjoint cycles (ignoring fixed points) in a permutation $\sigma_n \in S_n$ of maximum order $m_n$. If $k= c_n$ then there is a $k$-subset of $\{1,\ldots, n\}$ containing exactly one element from each cycle of $\sigma_n$. In the action of $\sigma_n$ on $k$-subsets, this subset is in a cycle of length $m_n$, the maximum possible. Multiplying $\sigma_n$ by a suitable element of the bottom group $H$ will give a cycle of length $2m_n$.

I still think it is a tricky problem to say much for general $k$.

I think this quickly leads to difficult questions.

Thinking of $B_n$ as the semidirect product $H \rtimes S_n$ where $H \cong C_2 \times \cdots \times C_2$, we can factorize each element of $B_n$ as $h t$ where $h \in H$ and $t \in S_n$. If $t$ has order $r$ then $(ht)^r = h'$ for some $h' \in H$, of order at most $2$. Choosing $h$ suitably, we can assume $h'$ is not the identity. Hence the maximum order of an element of $B_n$ is twice the maximum order in $S_n$, say $m_n$. Landau proved that

$$\log m_n \sim \sqrt{n \log n} \quad\text{as $n \rightarrow \infty$.}$$

For $k$ reasonably large (but less than $n$), my guess is that this means there is an element of $B_n$ having a cycle of length $2m_n$ in the action of $B_n$ on $k$-faces.

If we position the hypercube so that its $(n-1)$-faces are determined by the equations $x_i = 1$ or $x_i = -1$ for $i \in \{1,\ldots, n\}$ then the $k$-faces correspond to equations $x_{i_1} = \pm 1, \ldots, x_{i_{n-k}} = \pm 1$, with $n-k$ chosen signs. Hence the stabiliser of a $k$-face is a conjugate of $B_k \times S_{n-k}$. Therefore, restricting the action of $B_n$ on $k$-faces to the chosen top group $S_n$, we get $S_n$ acting on the cosets of $S_k \times S_{n-k}$, i.e. $S_n$ acting on $k$-subsets of $\{1,\ldots, n\}$. For this action, I would guess that for 'most' $k$, the maximum cycle length is $m_n$. As far as I know, no result of this form has been proved.

Edit Let $c_n$ be the number of distinct prime factors of $m_n$. In Proposition 7 in Ordre maximal d'un élément du groupe symmétrique, Bull. Soc. Math. France. 97 (1969) 129–191, Nicolas proved that

$$ c_n \sim 2 \sqrt{\frac{n}{\log n}}.$$

Note that $c_n$ is the number of disjoint cycles (ignoring fixed points) in a permutation $\sigma_n \in S_n$ of maximum order $m_n$. If $k= c_n$ then there is a $k$-subset of $\{1,\ldots, n\}$ containing exactly one element from each cycle of $\sigma_n$. In the action of $\sigma_n$ on $k$-subsets, this subset is in a cycle of length $m_n$, the maximum possible. Multiplying $\sigma_n$ by a suitable element of the bottom group $H$ will give a cycle of length $2m_n$.

I still think it is a tricky problem to say much for general $k$.

I think this quickly leads to difficult questions.

Thinking of $B_n$ as the semidirect product $H \rtimes S_n$ where $H \cong C_2 \times \cdots \times C_2$, we can factorize each element of $B_n$ as $h t$ where $h \in H$ and $t \in S_n$. If $t$ has order $r$ then $(ht)^r = h'$ for some $h' \in H$, of order at most $2$. Choosing $h$ suitably, we can assume $h'$ is not the identity. Hence the maximum order of an element of $B_n$ is twice the maximum order in $S_n$, say $m_n$. Landau proved that

$$\log m_n \sim \sqrt{n} \log n \quad\text{as $n \rightarrow \infty$.}$$

For $k$ reasonably large (but less than $n$), my guess is that this means there is an element of $B_n$ having a cycle of length $2m_n$ in the action of $B_n$ on $k$-faces.

If we position the hypercube so that its $(n-1)$-faces are determined by the equations $x_i = 1$ or $x_i = -1$ for $i \in \{1,\ldots, n\}$ then the $k$-faces correspond to equations $x_{i_1} = \pm 1, \ldots, x_{i_{n-k}} = \pm 1$, with $n-k$ chosen signs. Hence the stabiliser of a $k$-face is a conjugate of $B_k \times S_{n-k}$. Therefore, restricting the action of $B_n$ on $k$-faces to the chosen top group $S_n$, we get $S_n$ acting on the cosets of $S_k \times S_{n-k}$, i.e. $S_n$ acting on $k$-subsets of $\{1,\ldots, n\}$. For this action, I would guess that for 'most' $k$, the maximum cycle length is $m_n$. As far as I know, no result of this form has been proved.

I think this quickly leads to difficult questions.

Thinking of $B_n$ as the semidirect product $H \rtimes S_n$ where $H \cong C_2 \times \cdots \times C_2$, we can factorize each element of $B_n$ as $h t$ where $h \in H$ and $t \in S_n$. If $t$ has order $r$ then $(ht)^r = h'$ for some $h' \in H$, of order at most $2$. Choosing $h$ suitably, we can assume $h'$ is not the identity. Hence the maximum order of an element of $B_n$ is twice the maximum order in $S_n$, say $m_n$. Landau proved that

$$\log m_n \sim \sqrt{n} \log n \quad\text{as $n \rightarrow \infty$.}$$

For $k$ reasonably large (but less than $n$), my guess is that this means there is an element of $B_n$ having a cycle of length $2m_n$ in the action of $B_n$ on $k$-faces.

If we position the hypercube so that its $(n-1)$-faces are determined by the equations $x_i = 1$ or $x_i = -1$ for $i \in \{1,\ldots, n\}$ then the $k$-faces correspond to equations $x_{i_1} = \pm 1, \ldots, x_{i_{n-k}} = \pm 1$, with $n-k$ chosen signs. Hence the stabiliser of a $k$-face is a conjugate of $B_k \times S_{n-k}$. Therefore, restricting the action of $B_n$ on $k$-faces to the chosen top group $S_n$, we get $S_n$ acting on the cosets of $S_k \times S_{n-k}$, i.e. $S_n$ acting on $k$-subsets of $\{1,\ldots, n\}$. For this action, I would guess that for 'most' $k$, the maximum cycle length is $m_n$. As far as I know, no result of this form has been proved.

Edit Let $c_n$ be the number of distinct prime factors of $m_n$. In Proposition 7 in Ordre maximal d'un élément du groupe symmétrique, Bull. Soc. Math. France. 97 (1969) 129–191, Nicolas proved that

$$ c_n \sim 2 \sqrt{\frac{n}{\log n}}.$$

Note that $c_n$ is the number of disjoint cycles (ignoring fixed points) in a permutation $\sigma_n \in S_n$ of maximum order $m_n$. If $k= c_n$ then there is a $k$-subset of $\{1,\ldots, n\}$ containing exactly one element from each cycle of $\sigma_n$. In the action of $\sigma_n$ on $k$-subsets, this subset is in a cycle of length $m_n$, the maximum possible. Multiplying $\sigma_n$ by a suitable element of the bottom group $H$ will give a cycle of length $2m_n$.

I still think it is a tricky problem to say much for general $k$.