In the Lectures on n-Categories and Cohomology, Baez, Shulman, et al. "leave it as a puzzle" as to the nature of monoidal (-1)-categories. There are several ways to complete the table.
First, on nCat, they're listed as "trivial", just like the (-2)-category. This is unsatisfying, but I'll accept it: Somebody who groks the topic better has written this answer.
My next approach was decatigorification, going straight across from right to left on the table. A monoid (0-monoidal category) is a set (0-category) equipped with monoidal structure; there's a mapping which sends pairs of points (0-objects) to points, and a unit point, and associative laws. Decategorifying, a monoidal truth value is a truth value equipped with monoidal structure. But the (-1)-categories don't have (-1)-objects inside of them, do they?
But maybe we can find a simpler path. If we move diagonally-left on the table, then we preserve the rank of the category while altering how deep the arrows go. To copy the slogan and fill in the blanks, "we can think of a 0-category (a set) with just one j-morphism for j < 1 as a special sort of (-1)-category." There's only one nominee, the singleton set, so there should only be one monoidal truth value. This makes more sense.
Finally, we have the lecture notes, which do eventually answer the puzzle. They say, "a 0-category with one object is just a one-element set {x}, and hom(x, x) is just the equality x = x, which is 'True.'" I don't quite understand this explanation; how does equality manifest out of monoidal structure? I kind of understand the homotopy-based explanation, since equality/equivalence is central there, but not really.
Which of these lines of reasoning are decent and which are bogus? I hear that the entire concept of negative thinking is somewhat ill-conditioned; is this part of why? Or am I experiencing a Von Neumann "you don't understand things; you just get used to them" moment?
Given an $n$-category $\def\cC{\mathcal C}\cC$, they mention that $\def\Hom{\operatorname{Hom}}\Hom(x,y)$ is a $(n-j-1)$-category when $x$ and $y$ are parallel $j$-morphisms. By realising $n$-categories as "special $\infty$-categories", we are free to take $j$ as high as $j=n$ or even $j=n+1$, to allow us to deduce that
These are $0$-tuply monoidal categories, so how does the monoidal structure come into play? They arise when the parallel $j$-morphisms are taken more carefully. The key observations are:
Therefore, we obtain the pattern
So, if we want a $1$-tuply monoidal $(-1)$-category, we can take $n=j=k=1$ and see what happens: now we have a category $\cC$ and one parallel morphism $f:x\to y$ and we are looking at $\Hom(f,f)$. This is a $(-1)$-category, so it must be a truth value, but since $f=f$ this truth value must always be True. This is why a monoidal $(-1)$-category is trivial.
Regarding the monoidal structure on True, this has to be obtained by composition just like in all the other cases. "True" corresponds to the existence of the identity $2$-cell $\def\id{\operatorname{id}}\id_f:f=f$ in $\cC$, and the monoidal structure reflects the fact that $\id_f\circ\id_f=\id_f$. In terms of equations, this is saying we can compose $f=g$ with $g=h$ to get $f=h$, which is just transitivity of equality.
A similar way (which you remark as the easier way) of interpreting the pattern for $k$-tuply monoidal $n$-categories is to say that
(unwinding this, you will find that this says the exact same thing as the previous characterisation of these categories). With this in mind, a $1$-tuply monoidal $(-1)$-category ought to be a $0$-category with a single object; that is, a singleton set $\{x\}$. There is from here only one monoidal product which can be defined again ($x\circ x=x$), and we recover an object isomorphic to True with its unique monoidal structure.