Existence of a cts. function $f \geq 0$ satisfying $\int f(x)^n \, dx =1$ for all $n \geq 1$

No. Suppose such $f$ does exist to reach a contradiction. Then $\sup\limits_{x\in \mathbb R}f(x)=\lim\limits_{n\to\infty}\left(\int_{\mathbb R} f^n dL\right)^{1/n}=1$, so $f(x)\leq 1$ for all $x$. Hence $f-f^2$ is a nonnegative function, and by continuity $f-f^2$ is not always zero. This implies (again using continuity) that $\int_{\mathbb R} f-f^2 dL>0$.