The words $0,v_1,v_2,v_3,v_1+v_2,v_1+v_3,v_2+v_3,v_1+v_2+v_3$ form a group $C$ under addition. As observed by rschwieb, the given conditions imply that all the 8 vectors are distinct. We get six homomorphisms $f_i$ from $C$ to $\mathbb{Z}_2$ by mapping each vector to its bit at position $i$, $i=1,2,\ldots,6$. Because $|f_i(C)|$ is either 1 (if all the words have a zero at position $i$) or 2 (if both 1 and 0 occur at that position, we get that $|\mathrm{ker}(f_i)|$ is either 4 or 8. Therefore at each of the six bit positions we have a zero in either exactly 4 of the vectors of $C$, or in all 8 of them. Consequently, at each of the six bit positions we have a 1 in exactly four words, or a 0 in all 8 words.
Therefore the sum of the weights of the words of $C$ is divisible by four, and also $\le 4\cdot6=24$. OTOH the given inequalities imply that the sum of the weights is at least 23, so it must be exactly 24. Therefore the total "slack" (= the amount by which the inequality fails to be an equality) in these inequalities is exactly one.
We get a contradiction as follows. If all the vectors $v_i$ have an even weight, then so do the pairwise sums $v_i+v_j,i\neq j$. But the second set of inequalities then produce a total slack of three, which we showed to be impossible. So at least one of the vectors $v_i$ must have odd weight. In order for the first set of inequalities not to produce too much slack, exactly one of the vectors $v_i$, say $v_1$ must be of odd weight, and the other two must have weight four. But in this case $v_2+v_3$ has an even weight, and thus it will produce some slack in the second group of inequalities. Therefore the total slack would be at least two, which is also impossible.