*Note:* I'm not sure the construction I give is correct. There is an unresolved problem in proving it is a comonoid. See point 4 in the comonoid section. I don't have the time to resolve this right now.

I want to expand on Alexander Campbell's answer by giving the concrete promonoid as well as the necessary calculations. However my promonoid does not agree with the one he outlined. It's pretty late by now, so I hope I didn't miss anything.

# Conventions

Just to get everything straight, let $\mathcal V$ be a closed abelian category with tensor product $\otimes$ and unit $I$. $\mathcal C$ is the free $\mathcal V$-category with objects $\mathbb Z$, morphisms $\partial_n : n \to n - 1$ and relations $\partial_{n-1}\partial_n = 0$. A chain complex is a functor $C \to \mathcal V$.

A $\mathcal V$-profunctor $D \nrightarrow E$ is a $\mathcal V$-functor $E^{op} \otimes D \to \mathcal V$ (the swapped order of $D$ and $E$ is nLab convention). We pretend $\mathcal V\text -\mathbf{Prof}$ is a strict monoidal 2-category with monoid $\otimes$ and unit $\mathcal I$.

We identify objects with their identity arrows.

# Definition

We give all higher comultiplications of the comonoid.

For $n \in \{0,1,2,\ldots\}$ we define $(-|-)_n : \mathcal C \nrightarrow \mathcal C^{\otimes n}$, on objects as
$$
(k_1, \ldots, k_n| x)_n := \begin{cases}
I^{\oplus n} & \text{if } k_1 + \cdots + k_n = x + 1 \\
I & \text{if } k_1 + \cdots + k_n = x \\
0 & \text{otherwise.}
\end{cases}$$

For morphisms in each variable there is only one $\partial$ which has both domain and codomain non zero, in which case the domain is always $I$ and the codomain $I^{\oplus n}$. For the $j$th-contravariant variables the image of $\partial$ is then taken to be the $j$th-inclusion into the direct sum:
$$
(k_1,\ldots,\partial_{k_j + 1}, \ldots,k_n|k_1 + \cdots + k_n)_n := \iota_j.
$$

In the covariant variable the image of $\partial$ is the diagonal $I \to I^{\oplus n}$ but with signs:
$$
(k_1,\ldots, k_n| \partial_{k_1 + \cdots + k_n})_n := \iota_1 + (-1)^{k_1}\iota_2 + (-1)^{k_1 + k_2}\iota_3 + \cdots.
$$

We claim that $(-|-)_n$ is a comonoid in $\mathcal V\text-\mathbf{Prof}$ and the day convolution it defines is the tensor product of chain complexes.

# Tensor product of chain complexes

Consider chain complexes $C, D : \mathcal C \to \mathcal V$. We have
$$
C\otimes D := x \mapsto \int^{m,n} (m,n|x)\otimes Cm \otimes Cn.
$$

First we show that
$$
(C \otimes D)(x) = \bigoplus_{m + n = x} C(m)\otimes C(n)
$$
by showing the cones over the bases of these two colimits are the same.

Fix $x$ and consider $m,n$ with $m+n = x$. The following diagram is exemplary for the diagram defining the coend $C\otimes D$.

and here is the same diagram again:

Consider any cone over this. The component from the center term is already completely determined by the inclusions of its two summands on the left whose components are again determined by those of the top and bottom terms. Thus all components cones are given by those from terms of the form $(m,n|m+n)\otimes C(m) \otimes D(n)$, while nothing constrains these components. Thus the cones are in bijection with those over the base of $\bigoplus_{m+n=x} C(m) \otimes D(n)$ as claimed.

The image of $C\otimes D(\partial)$ is given at $C(m)\otimes D(n)$ by
$$
(m,n|\partial_{m+n})\otimes C(m) \otimes D(n) =
\begin{cases}
C(m) \otimes C(n) &\to (C(m)\otimes D(n))^{\oplus 2}\\
\iota_1 + (-1)^m\iota_2.
\end{cases}
$$
That is $\partial$ takes a term $C(m) \otimes D(n)$ at the top or bottom of the heart shaped diagram into one of the two term sums at the center of the heart shaped diagram. The two inclusions can then be pulled back and pushed along to the top and bottom term to yield the desired
$$
((C\otimes D)(\partial))_{m,n} = C(\partial_m)\otimes D(n) + (-1)^m C(m)\otimes D(\partial_n).
$$

# Comonoid in $\mathcal V\text-\mathbf{Prof}$

First note that $(-|-)_1$ is in fact $\mathcal C(-,-)$.

Let $j, k, n \in \{0,1,2,\ldots\}$ and sadly $k-j \leq 1$. We have
$$
(m_1,\ldots,m_n|x) \cong \int^y (m_1,\ldots,m_{j-1},y,m_{k+1},\ldots,m_n|x)\otimes(m_j,\ldots,m_k|y).
$$

Let $m_1,\ldots,m_n,x \in \mathbb Z$. We distinguish the following cases:

$m_1 + \cdots + m_n < x$: If we choose $y$ bigger than $m_j + \cdots + m_k$ the right term of the tensor product vanishes, otherwise the left term does.

$m_1 + \cdots + m_n = x$: The only non-zero term is $I\otimes I$.

$m_1 + \cdots + m_n = x + 1$: The coend is the colimit of the following diagram:

Which is the desired $I^{\oplus n}$

$m_1 + \cdots + m_n = x + 2$: Here the coend is the colimit of the following diagram:

If $k-j \leq 1$ this vanishes, but for $k-j >1$ I don't think it does. However since one could express any higher multiplication as one of arity 2 associated such that always $k-j < 1$ the term should also vanish for higher multiplications. So somewhere I must miss something, sadly I don't have the time to resolve this right now.

I think it might be possible to get rid of these terms by modifying the construction by taking $(m_1,\ldots,m_n,x)' := \mathcal C(m_1+\cdots+m_n,x)\otimes(m_1,\ldots,m_n,x)$. I think if we removed the condition that $\partial\partial = 0$ and also the signs we would get a tensor product of derivations where $\partial(x\otimes y) = \partial x \otimes y + x \otimes \partial y$, where these ghost terms would serve a purpose, as terms of degree $n$ could be constructed from terms of degrees higher than $n+1$. Tensoring with $\mathcal C(-,-)$ would then also ensure that $(-|-)_1 = \mathcal C(-,-)$ even if we do add relations to $\mathcal C$.

$m_1 + \cdots + m_n > x + 2$: As in 1. at least one side of the tensor product vanishes.

It remains to check naturality, which is straight forward keeping in mind the presentation of $(m_1,\ldots,m_n|m_1+\cdots+m_n-1)$ as the colimit of 3. above.