In this problem, $p$ must be greater than $1$. Here is a sketch of the proof. Let $\pi_{\leq N}$ be the coordinate projection on the subspace of $\ell_p$ spanned by $e_1, \dots, e_N$; let $\pi_{{}> N}$ be the coordinate projection on the closed subspace spanned by $e_{N+1}, e_{N+2}, \dots$.
Consider $f\in \ell_p^*$. Note that $f = f\pi_{{}> N} + f\pi_{{} \leq N}$. We have that $\|f\pi_{{}> N} \| \to 0$ as $N\to \infty$ since $\|f\pi_{{}> N}\| = \left(\sum_{i=N+1}^\infty |f_i|^q\right)^{1/q}$ (where $1/ p + 1/q = 1$).
Choose $N$ such that $\|f\pi_{{}> N} \|\leq \varepsilon$. Then, $\limsup_{n\to \infty} |f(u_n)| \leq \limsup_{n\to \infty} |f\pi_{{}\leq N}(u_n)| + \limsup_{n\to \infty} |f\pi_{{}> N}(u_n)| \leq 0 + \|f\pi_{{}> N}\|\cdot m \leq \varepsilon m.$
We get that for every $\varepsilon > 0$, $\limsup_{n\to \infty} |f(u_n)| \leq \varepsilon m$. Thus $\limsup_{n\to \infty} |f(u_n)| = 0$.
Here is a counterexample for $p=1$. Let $u_i = e_i$ and $f(u) = \sum_{i=1}^\infty u(i)$. The conditions of the problem hold but $f(u_i) = 1\not\to 0$.