You can form the $j$-fold self product, $A^j$, together with its addition morphism to $A$, $$\Sigma:A^j \to A.$$ Via Künneth, you have an inclusion $$\bigotimes_{i=1}^j H_1(A;\mathbb{Q}) \hookrightarrow H_j(A^j;\mathbb{Q}).$$ Now take the image of the cycle $\gamma_1\otimes \dots \otimes \gamma_j$ in $H_j(A^j;\mathbb{Q})$, and then take the pushforward via the morphism $\Sigma$ (proper and locally a fiber bundle) to get a class $\gamma\in H_j(A;\mathbb{Q})$. Similarly, given de Rham differentials, $$\omega_1,\dots,\omega_j \in H^{1,0}(A),$$
you can form $\omega= \omega_1\wedge \dots \wedge \omega_j$. Consider $\int_\gamma \omega$, or equivalently, the pairing, $$\int_\gamma \omega = \langle \Sigma^*(\omega_1\wedge \dots \wedge \omega_j),\gamma_1\otimes \dots \otimes \gamma_j\rangle.$$
Then you can ask whether or not, $$ \int_\gamma \omega = \text{det}\left[\int_{\gamma_\alpha} \omega_\beta\right]_{1\leq \alpha,\beta \leq j}?$$
Of course, ultimately, this has nothing to do with the differentials $\omega_\beta$ being contained in $H^{1,0}(A)$. Really this is a question about the singular cohomology of $A$: how does the Hopf algebra structure induced by addition behave? That question answers itself: the cohomology is a Hopf algebra. More precisely, it is the free exterior algebra on $H^1(A;\mathbb{Q})$ (with the usual trace on the top exterior power) equipped with its standard structure of cocommutative Hopf algebra. In particular, considered as an element in $H^1(A;\mathbb{Q})\otimes H^1(A;\mathbb{Q})$, $\Delta(\omega_\beta)$ equals $\omega_\beta\otimes 1 + 1\otimes \omega_\beta$, where $\Delta$ is pullback via addition, $$m:A\times A \to A,$$ together with the Künneth isomorphism, $$H^1(A\times A;\mathbb{Q}) = \left(H^1(A;\mathbb{Q})\otimes \mathbb{Q}\right) \oplus \left(\mathbb{Q}\otimes H^1(A;\mathbb{Q}) \right).$$ By my computation, this does imply that the Künneth component in $H^1(A;\mathbb{Q})\otimes \dots \otimes H^1(A;\mathbb{Q})$
of $\Sigma^*(\omega_1\wedge \dots \wedge \omega_j)$ is equal to $$\sum_{\sigma\in{\mathfrak{S}_j}} \text{sgn}(\sigma) \omega_{\sigma(1)}\otimes \dots \otimes \omega_{\sigma(j)}$$ (it would be wise for you to double-check that computation). Thus the pairing against $\gamma_1\otimes \dots \otimes \gamma_j$ equals $$\int_\gamma \omega = \sum_{\sigma\in{\mathfrak{S}_j}} \text{sgn}(\sigma) \langle \omega_{\sigma(1)}\otimes \dots \otimes \omega_{\sigma(j)}, \gamma_1\otimes \dots \otimes\gamma_j \rangle = \text{det}\left[\int_{\gamma_\alpha} \omega_\beta\right]_{1\leq \alpha,\beta \leq j}.$$
