Edit. After a conversation with Fedya I improve my previous answer. Let $f(x)=\sum_{k=1}^nA_n\cos nx$. If $f(x)\leq 1$, then $A_k\leq 2.$ This is the best possible estimate which holds for all $n$. For fixed $n$ it can be improved, but this is difficult.

Proof. We want to maximize $$A_k=\frac{2}{\pi}\int_0^\pi f(x)\cos kxdx$$ under the conditions that $$\int_0^\pi f(x)dx=0,\quad\mbox{and}\quad f(x)\leq 1.$$ It is clear that the maximizing function is $f^*(x)=1-\pi\delta$, where $\delta$ is the delta function representing the unit point mass at some point $x_0$ where takes its smallest value $\cos kx_0=-1$. And the $k$-th Fourier coefficient of $f^*$ is $2$.

Edit. After a conversation with Fedya I improve my previous answer. Let $f(x)=\sum_{k=1}^nA_n\cos nx$. If $f(x)\leq 1$, then $A_k\leq 2.$ This is the best possible estimate which holds for all $n$. For fixed $n$ it can be improved, but this is difficult.

Proof. We want to maximize $$A_k=\frac{2}{\pi}\int_0^\pi f(x)\cos kxdx$$ under the conditions that $$\int_0^\pi f(x)dx=0,\quad\mbox{and}\quad f(x)\leq 1.$$ It is clear that the maximizing function is $f^*(x)=1-\pi\delta$, where $\delta$ is the delta function representing the unit point mass at some point $x_0$ where takes its smallest value $\cos kx_0=-1$. And the $k$-th Fourier coefficient of $f^*$ is $2$.

Edit 2. Another version of this problem is obtained by replacing the restriction $f(x)\leq 1$ by the restriction $|f(x)|\leq 1$. In this case, the estimate will be $|A_n|\leq 4/\pi$, and this is also best possible in the class of all such trigonometric sums with any $n$.

This problem may be difficult, as stated. Let me show the best possible inequality $\max|A_1|<4/\pi$, where the maximum is taken over ALL trigonometric sums that you wrote (so that $n$ is not fixed.) Consider the polynomial $$P(z)=\sum_{1}^n A_nz^n;$$ This is a polynomial with real coefficients and your restriction is equivalent to saying that $|\mathrm{Re}\, P(z)|\leq 1$ for $|z|\leq 1$. Let us maximize $|A_1|$ in the class of ALL analytic functions in the unit disk, with real coefficients.

The extremal function is the conformal map of the unit disk onto the vertical strip $|\mathrm{Re}\, x|<1$. This is easily proved using Schwarz Lemma. The explicit form of this extremal function is $$f(z)=\frac{2i}{\pi}{\mathrm{Log}}\frac{1-iz}{1+iz}=\frac{4}{\pi}\left(z-z^3/3+z^5/5-z^7/7+\ldots\right).$$ This proves the inequality stated above. It is best possible, since this $f$ can be approximated by polynomials (of high degree).

Edit. After a conversation with Fedya I improve my previous answer. Let $f(x)=\sum_{k=1}^nA_n\cos nx$. If $f(x)\leq 1$, then $A_k\leq 2.$ This is the best possible estimate which holds for all $n$. For fixed $n$ it can be improved, but this is difficult.

Proof. We want to maximize $$A_k=\frac{2}{\pi}\int_0^\pi f(x)\cos kxdx$$ under the conditions that $$\int_0^\pi f(x)dx=0,\quad\mbox{and}\quad f(x)\leq 1.$$ It is clear that the maximizing function is $f^*(x)=1-\pi\delta$, where $\delta$ is the delta function representing the unit point mass at some point $x_0$ where takes its smallest value $\cos kx_0=-1$. And the $k$-th Fourier coefficient of $f^*$ is $2$.