Can $\nabla$ be called a "vector" in any meaningful way?

I used to think that $\nabla$ (or $\vec \nabla$) was just some fancy notation to represent some differential operators ($\nabla f \equiv \text{grad} \ f$, $\nabla \cdot \vec v \equiv \text{div} \ \vec v$, $\nabla \times \vec v \equiv \text{curl} \ \vec v$), which is particularly convenient because it happens to behave like a vector in algebraic manipulations.

However, I've read in some posts on this site (like this answer or this answer) that seem to suggest that there could in fact be a way to consider $\nabla$ as a vector in a formally meaningful way.

For example, quoting from this answer:

There are at least two layers of ideas here. First, as you say, the "dual space" $V^*$ to a real vector space is (by definition) the collection of linear maps/functionals $V\rightarrow \mathbb R$, with or without picking a basis. Nowadays, $V^*$ would more often be called simply the "dual space", rather than "covectors".

Next, the notion of "tangent space" to a smooth manifold, such as $\mathbb R^n$ itself, at a point, is (intuitively) the vector space of directional derivative operators (of smooth functions) at that point. So, on $\mathbb R^n$, at $0$ (or at any point, actually), $\{\partial/\partial x_1, \ldots, \partial/\partial x_n\}$ forms a basis for that vector space of directional-derivative operators.

It thus seems to me that there could be a meaningful, rigorous way to interpret $\nabla$ as an element of some vector space, maybe the dual space of an appropriate vector space of functions. Is this line of reasoning correct?

PS I am a physicist, not a mathematician, and I only have a very basic background on functional analysis and differential geometry.

In differential geometry, it is common to identify vectors (or "vector fields") and "derivations" (acting on the space of smooth functions). For instance, in $3$-dimensional space $\mathbb{R}^3$, the differential operator $$ \frac{\partial}{\partial x} + \frac{\partial}{\partial z} $$ is the same thing as the (constant) vector field $$ \overrightarrow{V} = (1, 0, 1).$$ This is because vector fields "act" on functions simply by declaring that $\overrightarrow{V}(f) := df(\overrightarrow{V})$ (what you might want to write $\overrightarrow{\nabla} f \cdot \overrightarrow{V}$). So, it is perfectly ok to write $\overrightarrow{V} = \frac{\partial}{\partial x} + \frac{\partial}{\partial z}$.

I wrote this because I thought you would be interested to know, but it does not really answer the question about $\nabla$. The symbol $\nabla$ is called a covariant derivative, and it is not a vector field (even though it kind of looks like one when using it on functions). This covariant derivative is another kind of differential operator, in general it is only well-defined when you have a Riemannian metric (which is the case on $\mathbb{R}^3$, the natural Euclidean metric). Contrary to the differential of a function, notions such as the gradient of a function or the divergence of a vector field require a metric. In $\mathbb{R}^3$, there is a last ingredient needed to give its full power to the notation $\nabla$: the cross product (which technically identifies vectors to bi-vectors, allowing to define the rotational as a vector field). These special features of $\mathbb{R}^3$ is the reason why the magical notations $$\overrightarrow{\nabla} f = \text{grad} \ f \qquad \overrightarrow{\nabla} \cdot \vec V = \text{div} \ \vec V \qquad \overrightarrow{\nabla} \times \vec V = \text{curl} \ \vec V$$ only work in $\mathbb{R}^3$, or I should say, only have partial generalizations to higher dimensions or more general spaces.

In conclusion, I would say that yes $\nabla$ is a real mathematical object that it is possible to define properly (it is more than just a notation), but no it is not quite right to say that it is a vector, and finally the formulas that you know involving it don't work 100% the same in general (that being said, it allows to do many other things).