Traces of powers of integral marices

Yes: for every $d\ge 3$ and all $A\in\mathrm{GL}_d(\mathbf{Z})$, there exists $n\in\{1,\dots,12^d\}$ such that the trace of $A^n$ is $\ge 3$. (This is probably far from sharp, and I don't know if we can do better than exponential.)

First part: I claim that if $t=(t_1,\dots,t_d)$ is a $d$-tuple in the unit circle, then there exists $n\in\{1,\dots,12^d\}$ such that all $t_1^n\dots t_d^n$ have argument in $[-\pi/6,\pi/6]$.

Indeed otherwise, denote $P_\theta$ as the set of $d$-tuples in the unit circle of argument in $\mathopen]-\theta,\theta\mathclose[$. Then for all $m\neq n$ with $m,n\in\{-12^d/2,\dots,12^d/2\}$, we have $t^{m-n}$ not in $P_{\pi/6}$. Define $V_n=t^nP_{\pi/12}$. Then the $V_i$, for $i\in\{-12^d/2,\dots,12^d/2\}$, are pairwise disjoint. If we endow the torus with the normalized Haar measure, the measure of $P_{\pi\theta}$, for $0\le\theta\le 1$ is equal to $\theta^d$. Hence the measure of $V_i$ is $(1/12)^d$. So $(12^d+1)/12^d\le 1$, that is, $1+12^{-d}\le 1$, contradiction.

I conclude with the following claim. Consider a matrix in $A=\mathrm{GL}_d(\mathbf{R})$ with $d\ge 3$, $|\det(A)|=1$, and all eigenvalues having argument in $[-\pi/6,\pi/6]$. Then the trace of $A$ is $>2$. In particular, if $A$ is integral then its trace is $\ge 3$.

Indeed, let $\lambda_1,\dots,\lambda_d$ be the eigenvalues, of modulus $r_1\ge r_2\ge \dots \ge r_d$ (we can suppose that conjugate eigenvalues are written consecutively). If $r_3\ge 1$, then the trace is $\ge (\sqrt{3}/2)(r_1+r_2+r_3)\ge 3\sqrt{3}/2> 2$.

Now suppose $r_3<1$. Note that $\prod r_i=1$, so $r_1r_2=\prod_{i\ge 3}r_i^{-1}>1$, so $r_2>1/r_1$. If $\lambda_1$ and $\lambda_2$ are real, then the trace is $>r_1+r_2>r_1+r_1^{-1}\ge 2$. If $\lambda_1$ is real and $\lambda_2$ is nonreal, then $r_2=r_3$ and hence the trace is $\ge r_1+2(\sqrt{3}/2)r_2>r_1+r_2>r_1+r_1^{-1}\ge 2$. If $\lambda_1$ is non-real and $\lambda_3$ is real, then $r_1=r_2$ and the trace is $>\sqrt{3}r_1+r_2\ge r_1+r_1^{-1}\ge 2$. Finally if both $\lambda_1$ and $\lambda_3$ are non-real, then $k\ge 4$, $r_1=r_2$, $r_3=r_4$ the trace is $\ge \sqrt{3}(r_1+r_3)$; we have $r_1^2r_3^2\ge 1$ implying $r_3\ge r_1^{-1}$ and hence the trace is $\ge \sqrt{3}(r_1+r_1^{-1})\ge 2\sqrt{3}>3$.

Thue's lemma allows to get an exponential bound for $k(n)$, though it seems to be not sharp. More specifically, idea is the following. We partition the unit circle onto three equal arcs $A,B,C$ and for each $k$ encode a sequence $(\lambda_1^k,\dots,\lambda_n^k)$ by a sequence of $n$ letters A,B,C which correspond to arcs containing arguments of $\lambda_i^k$. By pigeonhole principle two sequences concide for some positive integers $k,m$ not exceeding $3^n+1$. It means that for $h:=|k-m|$ real parts of all numbers $\lambda_i^h$ are positive and at least half of their absolute values, thus $\sum \lambda_i^h\geqslant \sum |\lambda_i|^h/2\geqslant n/2$.