In the case of a single variable, see for example this article concerning the limit $\lim_{n\to\infty}f^{(n)}(x)$ for a smooth function $f:\mathbb R\to\mathbb R$.
Also, if $f:\mathbb R\to\mathbb R$ is analytic, then for sufficiently small $h$ we have $e^{h\frac{d}{dx}}f(x)=f(x+h)$.
Just expand the exponential and you will see the Taylor series.
Thus a shift can be seen as an infinite order differential operation, but this only makes sense for analytic functions.
In your case one can also make sense of taking infinitely many derivatives, but one has to be careful in defining what everything means.
Here is an attempt:
Let $X_n$ denote the space of real sequences $(x_i)_{i\in\mathbb N}$ with $x_i=0$ when $i>n$, and let $X=\bigcup_{n\in\mathbb N}X_n$.
Let us call a function $f:X\to\mathbb R$ smooth if the restriction $f|_{X_n}$ is smooth for each $n$ in the usual sense.
Define $D^nf(x)=\frac{\partial^nf(x)}{\partial x_1\cdots\partial x_n}$.
Now it is reasonable to ask whether the limit $D^\infty f(x)=\lim_{n\to\infty}D^nf(x)$ exists.
For this limit to exist and be nonzero, we need to have $\lim_{n\to\infty}\partial_n\log(D^{n-1}f)=\lim_{n\to\infty}\frac{D^nf}{D^{n-1}f}=1$, so that in a way $f$ depends "asymptotically exponentially" on $x_n$.
There seem to be many smooth functions $f:X\to\mathbb R$ for which $D^\infty f$ exists. Take for example $f(x)=g(x_1,\dots,x_{17})\exp(\sum_{i=1}^\infty x_i)$ where $g$ is a smooth function.