Consider the union of intervals $[(2j+1) 2^{-n^2}, (2j+2) 2^{-n^2}]$, $j=0 \ldots 2^{2n-2}-1$, $n=1,2,\ldots$, and $\{0\}$.
EDIT: Let $I(j,n) = [(2j+1) 2^{-n^2}, (2j+2) 2^{-n^2}]$ and $U(N) = \bigcup_{n=N+1}^{\infty}\bigcup_{j=0}^{2^{2n-2}-1} I(j,n)$ Then $\widehat{\chi_{I(j,n)}}(2^{n^2} \pi) = 2^{1-n^2} i/\pi$. For $m < n$, $\widehat{\chi_{I(j,m)}}(2^{n^2} \pi) = 0$. On the other hand, $\left|\widehat{\chi_U(n)}(\xi)\right| \le m(U(n)) \le 2^{-n^2}$. So $\left|\widehat{\chi_K}(2^{n^2} \pi)\right|\cdot (2^{n^2} \pi) = \Omega(2^{n})$ as $n \to \infty$
EDIT: If you want a $K$ with empty interior, you can proceed as follows. Start with $K_0$ as above (which is the union of $\{0\}$ and intervals with rational endpoints) and a sequence $\xi_m$ such that $|\widehat{\chi_{K_0}}(\xi_m)| > m |\xi_m|$, and a sequence $\{r_n\}$ of irrationals dense in $K_0$. I'll choose rational numbers $a_n,b_n$ with $a_n < r_n < b_n$ and take $K_n = K_{n-1} \backslash (a_n, b_n)$, and then $K = \bigcup_n K_n$ will have empty interior.
By the fact that $r_n$ is irrational, either $r_n \notin K_{n-1}$ (in which case we take $a_n$ and $b_n$ so that $(a_n,b_n) \cup K_{n-1} = \emptyset$) or $r_n$ is in the interior of $K_{n-1}$, in which case we will want $[a_n,b_n] \subset K_{n-1}$. In the first case $\widehat{\chi_{K_n}} = \widehat{\chi_{K_{n-1}}}$, in the second $\widehat{\chi_{K_n}} = \widehat{\chi_{K_{n-1}}} - \widehat{\chi_{(a_n,b_n)}}$. Now $|\widehat{\chi_{(a_n,b_n)}}(\xi)| \le b_n - a_n$, so if $b_n - a_n$ is small enough and $|\widehat{\chi_{K_{n-1}}}(\xi_m) > (m/2) |\xi_m|^{-1}$ for $m = m_1, m_2, \ldots, m_{n-1}$, $|\widehat{\chi_{K_{n-1}}}(\xi_m)| > (m/2) |\xi_m|^{-1}$ for those same $m$, with $t_n < m_1$. Now since $|\widehat{\chi_{(a_n,b_n)}}(\xi)| = O(1/|\xi|)$, we can take $m_n$ to be any sufficiently large $m$ and we will have $|\widehat{\chi_{K_n}}(\xi_{m_n})| > (m_n/2) |\xi_{m_n}|^{-1}$. In the limit we will have $|\widehat{\chi_K}(\xi_{m_n})| \ge (m_n/2) |\xi_{m_n}|^{-1}$ for all $n$.