This definitely looks like an error in Grothendieck's paper! I think I have figured out what he meant. To simplify notation, I will concentrate on the case where Grothendieck builds his example: $H^3$ of a product of $3$ elliptic curves. I prefer to think in homology for this purpose, so I'll be talking about $H_3(E_1 \times E_2 \times E_3)$.
Put $X = E_1 \times E_2 \times E_3$ where $E_i = \mathbb{C}/(\mathbb{Z}+\mathbb{Z} \tau_i)$. I'll write $z_i$ for the coordinate on the universal cover of $E_i$, so $dz_i$ is the nonvanishing holomorphic $(1,0)$-form on $E_i$.
Let $V$ be the subspace of $H_3(X, \mathbb{Q})$ spanned by the images of all $H_3(Y, \mathbb{Q})$, as $Y$ ranges over $2$-dimensional subvarieties of $X$. Hodge's "general conjecture" is an attempted description of $V$. (The famous Hodge conjecture deals with $H_{2 \dim Y}(Y)$, so we are talking about the span of the fundamental classes; this less famous one allows the subscript on homology to be independent of the dimension of $Y$.)
Specifically, Hodge conjectures that $\alpha \in H_3(X, \mathbb{Q})$ lies in $V$ if and only if $\int_{\alpha} \omega=0$ for any $\omega \in H^{(3,0)}(X)$. In fact, $H^{(3,0)}(X)$ is one dimensional, spanned by $d z_1 \wedge d z_2 \wedge dz_3$, so this criterion says that $\int_{\alpha} d z_1 \wedge d z_2 \wedge dz_3=0$. Let's denote the space of $\alpha$ obeying this condition by $V'$. It is easy to see that $V \subseteq V'$, since $ d z_1 \wedge d z_2 \wedge dz_3|_Y=0$ for any algebraic surface $Y$.
Grothendieck observes that, for "trivial reasons", $V$ is even dimensional. However, he shows that $V'$ need not be.
By Kunneth, we have
$$H_3(X, \mathbb{Q}) = \bigoplus_{i_1+i_2+i_3=3} H_{i_1}(E_1, \mathbb{Q}) \otimes H_{i_2}(E_2, \mathbb{Q}) \otimes H_{i_3}(E_3, \mathbb{Q}).$$
Every summand other than $H_1 \otimes H_1 \otimes H_1$ is easily seen to be in $V$. So it is natural to consider the inclusion $V \cap (H_1\otimes H_1 \otimes H_1) \subseteq V' \cap (H_1\otimes H_1 \otimes H_1)$, which will be equality if and only if $V \subseteq V'$ is equality. I believe that Grothendieck, without bothering to say so, has switched to working with these subspaces in his dimension computation. Each of the other summands is two-dimensional, so this doesn't effect the parity of the dimension.
So, let's compute the dimension of $V' \cap (H_1\otimes H_1 \otimes H_1)$. Let $\gamma^0_i$ be the $1$-cycle $\mathbb{R}/\mathbb{Z}$ in $E_i$, and let $\gamma^1_i$ be the $1$-cycle $\mathbb{R} \tau_i/\mathbb{Z} \tau_i$. So $\gamma^0_i$, $\gamma^1_i$ is a basis of $H_1(E, \mathbb{Q})$. So a basis for $H_1 \otimes H_1 \otimes H_1$ is the $8$ products $\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}$, where each $j_i$ is $0$ or $1$. We have $\int_{\gamma_i^j} dz_i = \tau_i^j$, so
$$\int_{\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}} d z_1 \wedge d z_2 \wedge dz_3 = \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}.$$
Putting
$$\alpha = \sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3)\ [\gamma_1^{j_1} \times \gamma_2^{j_2} \times \gamma_3^{j_3}] \in H_3(X, \mathbb{Q}),$$
we see that $\alpha \in V'$ if and only if
$$\sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3) \ \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}=0.$$
Thus, $V' \cap (H_1\otimes H_1 \otimes H_1)$ is identified with the $\mathbb{Q}$-vector space
$$\left\{ c \in \mathbb{Q}^{\{0,1\}^3} : \sum_{j_1, j_2, j_3 =0}^1 c(j_1, j_2, j_3) \ \tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}=0 \right\}.$$
We see that the dimension of $V' \cap (H_1\otimes H_1 \otimes H_1)$ is $2^3-N$ where $N$ is the dimension of the $\mathbb{Q}$-vector space spanned by the monomials $\tau_1^{j_1} \tau_2^{j_2} \tau_3^{j_3}$.
As Grothendieck points out, if $\tau_1=\tau_2=\tau_3$ is a cubic irrational, these monomials span a $3$-dimensional vector space. As you say, for generic $\tau_i$, the monomomials span an $8$ dimensional vector space, so $V' \cap (H_1\otimes H_1 \otimes H_1)=0$. This reflects that the obvious classes, coming from $H_3(E_i \times E_j)$ with $1 \leq i < j \leq 3$, are all in the other summands of $H_3(X)$.
