= Solution
Use the periodic <modulus of continuity>
$$
\omega(f,\delta)=\sup_{|h|\leq\delta}\sup_x|f(x+h)-f(x)|.
$$
Subdividing a displacement $t$ into at most $\lceil|t|/\delta\rceil$ steps of size at most $\delta$ proves the chaining inequality
$$
|f(x-t)-f(x)|\leq\left(1+\frac{|t|}{\delta}\right)\omega(f,\delta).
$$
For $|t|\leq\delta$, the sharper bound $\omega(f,\delta)$ holds directly.
Set $\delta=n^{-\alpha}$, where $0<\alpha<1$ and $n\geq1$. Positivity and unit normalized mass of the <Fejér kernel> give
$$
|\sigma_n(f;x)-f(x)|\leq\frac1\pi\int_{-\pi}^{\pi}|f(x-t)-f(x)|F_n(t)\,dt.
$$
The portion $|t|\leq\delta$ is at most $\omega(f,\delta)$. For $0<|t|\leq\pi$, $\sin(|t|/2)\geq |t|/\pi$ and $|\sin(nt/2)|\leq1$, so
$$
F_n(t)\leq\frac{\pi^2}{2nt^2}.
$$
Apply this bound and the chaining inequality to the two tails:
$$
\begin{aligned}
\|\sigma_nf-f\|_\infty
&\leq\omega(f,\delta)\left[1+\frac\pi n\int_\delta^\pi\left(1+\frac t\delta\right)\frac{dt}{t^2}\right]\\
&=\omega(f,\delta)\left[1+\frac\pi n\left(\frac1\delta-\frac1\pi+\frac1\delta\log\frac\pi\delta\right)\right]\\
&\leq\omega(f,\delta)\left[1+\pi n^{\alpha-1}\bigl(1+\log\pi+\alpha\log n\bigr)\right].
\end{aligned}
$$
Let $\beta=1-\alpha>0$. Calculus gives $\sup_{u\geq1}u^{-\beta}\log u=1/(e\beta)$, attained at $u=e^{1/\beta}$, and $u^{-\beta}\leq1$. Thus a possible explicit constant for the <fractional-scale Fejér approximation bound> is
$$
c_\alpha=1+\pi\left(1+\log\pi+\frac{\alpha}{e(1-\alpha)}\right),
\qquad\boxed{\|\sigma_nf-f\|_\infty\leq c_\alpha\omega(f,n^{-\alpha}).}
$$
The constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic <function> is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha<1$; a crude global-oscillation bound would not prove the whole requested range.
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