The argument is a series of more and more general cases.

(1) The map $H_{DR}^1(\mathbb{C}^{\ast}) \otimes \mathbb{C} \to H_B^1(\mathbb{C}^{\ast}) \otimes \mathbb{C}$ is multiplication by $(2 \pi i) \mathbb{Q}$. Proof: $H^1_{DR}$ is generated by $dz/z$. Let $\gamma$ be a closed path in $\mathbb{C}^{\ast}$ which circles the origin once, then $H^1_{B}$ is generated by $[\gamma] \mapsto 1$. Since $\int_{\gamma} dz/z= 2 \pi i$, we see that $H^1_{DR} \otimes \mathbb{C} \to H^1_B \otimes \mathbb{C}$ is multiplication by $2 \pi i$ with respect to the given generators.

(2) $H_{DR}^j(\mathbb{C}^k \times (\mathbb{C}^{\ast})^{\ell}) \otimes \mathbb{C} \to H_B^j(\mathbb{C}^k \times (\mathbb{C}^{\ast})^{\ell}) \otimes \mathbb{C}$ is multiplication by $(2 \pi i)^j \mathbb{Q}$. Proof: The Kunneth isomorphism is compatible with everything in the picture.

(3) $H^{2j}_{DR}(\mathbb{P}^n) \otimes \mathbb{C} \to H^{2j}_{B}(\mathbb{P}^n) \otimes \mathbb{C}$ is multiplication by $(2 \pi i)^j \mathbb{Q}$. Proof: Write down an open cover of $\mathbb{P}^n$ where every intersection of open sets is of the form (2). We get corresponding spectral sequence in $H_{DR}$ and $H_B$, and compatible maps between them. Keep track of degrees to get the correct power of $2 \pi i$.

(4) Let $X$ be smooth projective of complex dimension $n$. Then $H^{2n}_{DR}(X) \to H^{2n}_B(X)$ is multiplication by $(2 \pi i)^n \mathbb{Q}$. Proof: Write $X$ as a finite cover of $\mathbb{P}^n$ and draw the corresponding commuting square in $H^{2n}$.

(5) (The one that involves determinants.) Let $X$ be smooth projective of dimension $n$. Let $\zeta_B$ be the singular cohomology class of a hyperplane and let $\zeta_{DR}$ be the deRham class $(2 \pi i) \zeta_B$. Fix a cohomological degree $k \leq n$; the claim for $k$ larger than $n$ follows by Poincare duality.

Hard Lefschetz tells us that
$$(\alpha, \beta) \mapsto \alpha \cup \beta \cup \zeta_T^{n-k}$$
is a perfect pairing $H^k_T \times H^k_T \to H^{2n}_T$, where $T$ is either $DR$ or $B$. We instead write this as a pairing $H^k_T \times H^k_T \to \mathbb{Q}$ given by
$$\langle \alpha, \beta \rangle_T := (\alpha \cup \beta \cup \zeta_T^{n-k})/\zeta_T^n.$$
Division by $\zeta_T^n$ makes sense because $H^{2n}$ is one dimensional.

Observe that we have
$$\langle \alpha, \beta \rangle_{DR} = \langle \alpha, \beta \rangle_{B} (2 \pi i)^{-k}$$
as pairings on $H^k_{DR} \otimes \mathbb{C} \cong H^k_B \otimes \mathbb{C}$, because of the distinction between $\zeta_{DR}$ and $\zeta_T$.

Choose bases for these cohomology theories and let $A_{DR}$ and $A_B$ be the matrices of the Hard-Lefschetz pairing in these two bases. Let $P$ be the matrix of the period isomorphism $H_{DR}^K \to H_B^k$. The above discussion turns into the matrix equation
$$A_{DR} = P^T A_B P (2 \pi i)^{-k}.$$
So $(\det A_{DR})= (\det A_B) (\det P)^2 (2 \pi i)^{-k b_k}$ where $b_k = \dim H^k$. The two $A$-matrices have rational entries and, since the pairings were perfect, $\det A_B$ and $\det A_{DR} \neq 0$. So $(\det P)^2 \in (2 \pi i)^{k b_k} \mathbb{Q}^{\times}$ as desired.