Let $v(\beta) := \sum\_{n\le X} e(n\beta)$ where $e(\alpha) := e^{2\pi i \alpha}$. It is not hard to show that 
$$\log X\ll \int\_0^1 |v(\beta)| d\beta\ll \log X$$
and by considering the underlying Diophantine equation, we have that for $k\ge 0, $
$$\int\_0^1 |v(\beta)|^{2k}d\beta\sim C\_kX^{2k - 1}.$$


For general $p\ge 0$, is the order of magnitude of $\int\_0^1 |v(\beta)|^pd\beta$ known? 


Edit: corrected the actual exponent for the $2k$th moment.



Moments of certain exponential sum
Yes. Without loss of generality, $X\geq 2$ is an integer. Then we have explicitly
$$ v(\beta)=e\left(\frac{(X+1)\beta}{2}\right)\frac{\sin(\pi X\beta)}{\sin(\pi\beta)},\qquad\beta\not\in\mathbb{Z}.$$
It follows for any $p>0$ that
$$\int\_{|\beta|\leq\frac{1}{4X}}|v(\beta)|^p\,d\beta\asymp\_p\int\_{|\beta|\leq \frac{1}{4X}}X^p\,d\beta\asymp\_p X^{p-1}.$$
Now we estimate the rest of the integral carefully.
On the one hand,
$$\int\_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta\ll\_p\int\_{\frac{1}{4X}<|\beta|\leq \frac{1}{2}}|\beta|^{-p}\,d\beta\asymp\_p \begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$
On the other hand,
\begin{align\*}\int\_{\frac{1}{4X}<|\beta|\leq\frac{1}{2}}|v(\beta)|^p\,d\beta&\gg\_p\sum\_{1\leq m\leq\frac{X}{2}}\int\_{\frac{1}{4}<m-|X\beta|<\frac{3}{4}}\frac{d\beta}{|\beta|^{p}}\\
&\asymp\_p\sum\_{1\leq m\leq\frac{X}{2}}\frac{1}{X}\left(\frac{X}{m}\right)^p\asymp\_p\begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}\end{align\*}
Putting together these three bounds,
$$\int\_{|\beta|\leq 1/2}|v(\beta)|^p\,d\beta\asymp\_p
\begin{cases}X^{p-1},&p>1;\\\log X,&p=1;\\1,&p<1.\end{cases}$$
This also means that in your second display the correct exponent of $X$ is $2k-1$, not $k$.


