Here are three sets of data:
A B C
0 0 4.84
2 0 4.28
1 1.73 2.74
You can check that the (product moment) correlation coefficient between A and B is zero, whereas the correlation between B and C, and between A and C is 0.71. I observe that the correlation between A-C and B-C is close to $1/\sqrt{2}$, and experimenting this seems to be the "best" possible. To make that precise, the product of the correlations between A-C and B-C seems always to be less than 0.5.
This answers my question for three items per data set, however:
is the product of the non-zero correlations always less than 0.5 ?
Can similar collections of data be found with more than three items?
Pearson's (product moment) correlation coefficient may be defined for nonconstant samples $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$ by taking an inner product of the vectors with their mean values subtracted from each entry and "normalized" to have unit length.
To see this, start with the following "standard" definition (which can be rearranged in many ways equivalent in exact arithmetic):
$$ r_{xy} = \frac{\sum_{i=1}^n (x_i - \bar{x}) (y_i - \bar{y}) }{ \sqrt{\sum_{i=1}^n (x_i - \bar{x})^2)} \sqrt{\sum_{i=1}^n (y_i - \bar{y})^2}} $$
where $\bar{x},\bar{y}$ are the respective sample means.
Notice that if $u = (x_1-\bar{x},\ldots,x_n-\bar{x})$ and $v = (y_1-\bar{y},\ldots,y_n-\bar{y})$, then the above can be rewritten:
$$ r_{xy} = \frac{u}{||u||} \cdot \frac{v}{||v||} $$
where $||u||,||v||$ denote the usual Euclidean lengths of $u,v$.
In other words, by virtue of this interpretation as an inner product of unit vectors, the correlation coefficient $r_{xy}$ is the cosine of the angle between the (nonzero) vectors $u,v$. [Note that the assumption that $x,y$ were nonconstant samples implies $u,v$ are nonzero vectors.]
In particular $r_{xy} = 0$ occurs precisely when vectors $u,v$ are orthogonal to each other.
In the case $n=3$ (the three-item case mentioned in the Question), the vectors $u,v$ are constrained to lie in a two-dimensional subspace, the plane of vectors whose coordinates sum to zero (because we subtracted the mean from $x,y$ respectively).
Now let's introduce an equally sized third nonconstant sample $z = (z_1,\ldots,z_n)$ and its mean-subtracted vector $w = (z_1-\bar{z},\ldots,z_n)$. Then the additional correlation coefficients are similarly expressed by inner products of unit vectors:
$$ r_{xz} = \frac{u}{||u||} \cdot \frac{w}{||w||} $$
$$ r_{yz} = \frac{v}{||v||} \cdot \frac{w}{||w||} $$
Like vectors $u,v$, $w$ is constrained to the subspace of vectors whose coordinates sum to zero. Again, if $n=3$ this is just a plane.
If we assume $u,v$ are orthogonal, then we may without loss of generality assume that $u/||u||$ is positioned as the $X$-axis and $v/||v||$ as the $Y$-axis of the plane. The third unit vector $w/||w||$ could be anywhere in the same plane, and we can ask what maximum value can be attained by $ r_{xz} r_{yz} $.
Let us parameterize the possible locations of $w/||w||$ by $(\cos \theta,\sin \theta)$. Then using the previous observation about the correlation coefficient being the cosine of the angle between respective unit vectors:
$$ r_{xz} r_{yz} = \cos \theta \sin \theta = \frac12 \sin 2 \theta $$
Since the maximum value of sine is $1$, this shows the maximum product of the correlation coefficients is $1/2$, so never more than $0.5$ as the first part of your Question asks.
The second part, where we allow more than three items in each sample, will have the same conclusion: the product is never more than $0.5$. Geometrically the reason is that vectors $u,v$ always determine some plane in which both of them lie. The inner product of unit vector $w/||w||$ with each of $u/||u||$ and $v/||v||$ is preserved by projecting it orthogonally on the that plane. This can shorten the vector (the projection will be less than a unit in length unless it already lies in the plane), but cannot change the inner products.
Thus even if the sample size $n \gt 3$, we still cannot have the product $r_{xz} r_{yz}$ more than $0.5$.