I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An Introduction) which is one place to understand the subtleties of this issue.

One thing I will mention is that the Sobolev embedding theorem implies sufficient conditions for Holder continuity. If, for example, $n^2 \hat{f}(n)$ is summable ($f \in H^1$), then $f$ is $C^{0,\alpha}$ for $\alpha<\frac{1}{2}$. More generally, you can find conditions based on the following idea:

$|f|_\alpha \leq \sum_n |\hat{f}(n)| |n|^\alpha = \sum_n |n|^{\alpha+\frac{1}{2}+\epsilon}|\hat{f}(n)| \frac{1}{|n|^{\frac{1}{2}+\epsilon}} \leq \sum_n |n|^{2\alpha+1+2\epsilon}|\hat{f}(n)|^2 \sum_n \frac{1}{|n|^{1+2\epsilon}}$

Therefore if $|n|^{2\alpha+1}\hat{f}(n)$ is summable then $f$ is Holder continuous of any order strictly less than $\alpha$.

I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An Introduction) which is one place to understand the subtleties of this issue.

One thing I will mention is that the Sobolev embedding theorem implies sufficient conditions for Holder continuity. If, for example, $n^2 |\hat{f}(n)|^2$ is summable ($f \in H^1$), then $f$ is $C^{0,\alpha}$ for $\alpha<\frac{1}{2}$. More generally, you can find conditions based on the following idea:

$|f|_\alpha \leq \sum_n |\hat{f}(n)| |n|^\alpha = \sum_n |n|^{\alpha+\frac{1}{2}+\epsilon}|\hat{f}(n)| \frac{1}{|n|^{\frac{1}{2}+\epsilon}} \leq \sum_n |n|^{2\alpha+1+2\epsilon}|\hat{f}(n)|^2 \sum_n \frac{1}{|n|^{1+2\epsilon}}$

Therefore if $|n|^{2\alpha+1}|\hat{f}(n)|^2$ is summable then $f$ is Holder continuous of any order strictly less than $\alpha$.