This is false; the result is not perfectly analogous because the definition of the spectral radius contains a maximum. Here is a counterexample.

Take $$ A_1 = \begin{bmatrix}2 & 0\\ 0 & 1\end{bmatrix}, \quad A_2 = \begin{bmatrix}1 & 0\\ 0 & 2\end{bmatrix}. $$ Then, each product $A_{j_n}\dotsm A_{j_1}$ of length $n$ contains at least $n/2$ elements equal to $A_1$ or $n/2$ elements equal to $A_2$, and thus it has spectral radius at least $2^{n/2}$. This proves that $LSR(\mathcal{A}) \geq 2^{1/2}$. On the other hand, $LSR(\mathcal{B}) = LSR(\mathcal{C}) = 1$, clearly.

If you replace $\rho$ with a function that computes the minimum modulus of an eigenvalue of a matrix, then the corresponding result should hold.

This is false; the result is not perfectly analogous because the definition of the spectral radius contains a maximum. Here is a counterexample.

Take $$ A_1 = \begin{bmatrix}2 & 0\\ 0 & 1\end{bmatrix}, \quad A_2 = \begin{bmatrix}1 & 0\\ 0 & 2\end{bmatrix}. $$ Then, each product $A_{j_n}\dotsm A_{j_1}$ of length $n$ contains at least $n/2$ elements equal to $A_1$ or $n/2$ elements equal to $A_2$, and thus it has a diagonal element at least as large as $2^{n/2}$, and $\rho(A_{j_n}\dotsm A_{j_1}) \geq 2^{n/2}$. This proves that $LSR(\mathcal{A}) \geq 2^{1/2}$. On the other hand, $LSR(\mathcal{B}) = LSR(\mathcal{C}) = 1$, clearly.

If you replace $\rho$ with a function that computes the minimum modulus of an eigenvalue of a matrix, then the corresponding result should hold.