Let $1\leq p\leq\infty$. For each $n\in\mathbb{N}$ we denote by $(e_k^{(n)})_{k=1}^n$ the canonical basis for $\ell_p^n$, and fix some other basis $(x_k^{(n)})_{k=1}^n$ for a given $n$-dimensional Banach space $X_n$. Assume that each $(x_k^{(n)})_{k=1}^n$ is $K$-unconditional for some fixed $K\in[1,\infty)$.
Goal. We want to show that $(x_k^{(n)})_{k=1}^\infty$ are uniformly equivalent to $(e_k^{(n)})_{k=1}^n$. More precisely, we want to find $C\in[1,\infty)$ such that for every $n\in\mathbb{N}$ and every scalar sequence $(a_k)_{k=1}^n$, \begin{equation}\tag{1}\frac{1}{C}\|\sum_{k=1}^na_ke_k^{(n)}\|_{\ell_p^n}\leq\|\sum_{k=1}^na_kx_k^{(n)}\|_{X_n}\leq C\|\sum_{k=1}^na_ke_k^{(n)}\|_{\ell_p^n}.\end{equation}
Question. Are there any easy (or relatively easy) shortcuts for achieving the above goal?
Discussion. Obviously we may assume $a_k\geq 0$.
In case $p=1$, the right-hand estimate in $(1)$ follows from the triangle inequality so that we only need to show the left-hand estimate.
The case $p=\infty$ is even easier as we always get the left-hand inequality in $(1)$ by the weakness of the $\infty$-norm, and, thanks to uniform unconditionality, the right-hand inequality in $(1)$ follows as long as $\|x_1^{(n)}+\cdots+x_n^{(n)}\|_{X_n}\leq M$ for some $M\in[1,\infty)$ chosen independently of $n$. (See this question for a short explanation of this.)
However, I don't see any easy methods for the cases $1<p<\infty$. Am I out of luck? What if, say, each $(x_k^{(n)})_{k=1}^n$ was uniformly equivalent to all its permutations? Or is there some other property which could make it easier?
Thanks guys!