On the cubic generalization $(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = v_1^3+v_2^3+v_3^3+v_4^3$ for the Euler four-square We are familiar with the Euler Four-Square identity,

$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2 ) = u_1^2+u_2^2+u_3^2+u_4^2$$

where,
$$u_1 = ae-bf-cg-dh\\
u_2 = af+be+ch-dg\\
u_3 = ag-bh+ce+df\\
u_4 = ah+bg-cf+de$$
or the product of two sums of four squares is itself a sum of four squares.

Tinkering about, I came across a cubic version,

$$(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = 6^{-3} (w_1^3+w_2^3+w_3^3+w_4^3)$$

where,
$$w_1= 9 + a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 + h^3\\
w_2= -9 - a^3 - b^3 - c^3 - d^3 + e^3 + f^3 + g^3 + h^3\\
w_3= 9 - a^3 - b^3 - c^3 - d^3 - e^3 - f^3 - g^3 - h^3\\
w_4= -9 + a^3 + b^3 + c^3 + d^3 - e^3 - f^3 - g^3 - h^3$$
Q: Does the cubic version have any number theoretic implications, like the set of sums of four cubes is closed under multiplication? Or is it just an interesting curiosity?

Summarizing some comments and adding a bit on top:
The identity is a special case of
$$24ABC = (\underbrace{A+B+C}_{w_1})^3 + (\underbrace{-A+B-C}_{w_2})^3
        + (\underbrace{-A-B+C}_{w_3})^3 + (\underbrace{A-B-C}_{w_4})^3
\tag{1}$$
where setting $C=9$ turns the left-hand side into $6^3AB$.
The settings
$$\begin{align}
    A &= a^3 + b^3 + c^3 + d^3
\\  B &= e^3 + f^3 + g^3 + h^3
\end{align}$$
are not necessary for $(1)$ but lead to an interesting specialization.
To make the right-hand side of $(1)$ a symbolic integer multiple of $6^3$,
we might want to require that each $w_i$ is divisible by $6$.
Straightforward calculations show that this happens if and only if both $A$
and $B$ are divisible by $3$ and $A+B$ is odd.
In other words,
$$\{A,B\}=\{6m,6n+3\}\quad\text{for some}\quad m,n\in\mathbb{Z}$$
Conversely, for every choice of $m,n\in\mathbb{Z}$ there exist
representations of $A$ and $B$ as sums of four cubes, such as
$$\begin{align}
     6m &= (m+1)^3 + (m-1)^3 + (-m)^3 + (-m)^3                   \tag{2}
\\   6n+3 &= n^3 + (-n + 4)^3 + (2n - 5)^3 + (-2n + 4)^3         \tag{3}
\end{align}$$
taken from the Alpertron
hyperlinked in Dietrich Burde's comment.
Specializing your identity to that scenario gives
$$18\,m\,(2n+1) =
(2 + m + n)^3 + (-1 - m + n)^3 + (1 - m - n)^3 + (-2 + m - n)^3 \tag{4}$$
which may give much smaller solutions for representations of $18q$ than $(2)$,
particularly if $q$ has an odd divisor near $\sqrt{2|q|}$ which we can use for
$2n+1$.
Examples:
$$\begin{array}{r|rr|rrr}
    18q & (2)\quad\text{with}               & m
        & (4)\quad\text{with}               & m,    & n
\\\hline
    18  & 4^3 + 2^3 + (-3)^3 + (-3)^3       & 3
        & 3^3 + (-2)^3 + 0^3 + (-1)^3       & 1     & 0
\\  504 & 85^3 + 83^3 + (-84)^3 + (-84)^3   & 84
        & 9^3 + (-2)^3 + (-6)^3 + (-1)^3    & 4     & 3
\end{array}$$
The following points may be worth mentioning:


*

*Replacing $2$ with $0$ in the right-hand side of $(4)$ turns the left-hand
side into $6\,m\,(2n-1)$ which is more widely applicable.

*Replacing $2$ with $1$ in the right-hand side of $(4)$ turns the left-hand
side into $24mn$, an easy specialization of $(1)$.

*$(1)$ itself, where applicable, might give solutions
with even smaller absolute maximum of the numbers involved, particularly if
$A,B,C$ can be chosen to be pairwise close.

