I am interested in computing the algebraic parts of $L(f, n, \chi)$ for a primitive form $f \in S_k(\Gamma_0(N))$ with wieght $k > 2$ twisted by a primitive Dirichlet character $\chi$ of conductor $m$. It is known that there are some non-zero complex numbers $\Omega^{\pm}$ depending on the sign of $\chi$ such that for any positive integer $n$ with $1 \le n \le k-1$, \begin{equation*} \Lambda(f, n, \chi) := \frac{c(m, n, \chi)}{\Omega^{\pm}}L(f, n, \chi) \in \mathbb{Z}[\chi], \end{equation*} for some constants $c(m,n,\chi) \in \mathbb{C}$ depending on $m, n \text{ and } \chi$ and $\mathbb{Z}[\chi]$ is the ring of integers generated by the values of $\chi$.
For $k = 2$, it is well known to compute $\Omega^{\pm}$. For $k > 2$, For a given such $f$ and $\chi$, I can compute the values of $L(f, n, \chi)$ with the Dokchitser's calculator as well as $c(m,n,\chi)$ but can not find a way to compute $\Omega^{\pm}$.
Question: Can anyone let me know how to compute $\Omega^{\pm}$ directly? A reference for this computation is also welcome.
Update #1: Following Chapter 1. on "On $p$-adic analogues of the conjecture of Birch and Swinnerton-Dyer" by Mazur, Tate and Teitelbaum, we have \begin{equation*} c(m, n, \chi) = \frac{(n-1)!m^n}{(-2\pi i)^{n-1}\tau(\chi)}, \end{equation*} where $\tau(\chi)$ is the Gauss sum of $\chi$. For the computations of $L(f, n, \chi)$ one way is to use the functional equation and evaluate the series of integrals involving the Fourier coefficients of $f$ within some precision (This is exactly how Dokchitser built in his L-value calculator). In Mazur, Tate and Teitelbaum, they presented some numerical values of $\Lambda(f, n, \chi)$ on page 11 for quadratic twists of the primitive form $f \in S_6(\Gamma_0(3))$. From there, I can obtain $\Omega^{+}$ with the values of $L(f, n, \chi)$ computed by Dokchitser's $L$-function calculator.
I hope this update can explain it more explicitly as reuns wants.
Update #2: As David pointed out, the form $f$ should be assumed that its Fourier coefficients are in $\mathbb{Q}$.