Is Del (or Nabla) an operator or a vector?

First, let's say that $\nabla$ and $\vec \nabla$ are two equivalent notations for the same "object". This notation is used in the representation of three important vector operators: gradient, curl and divergence.

The gradient operator acts on a scalar differentiable function $f(\vec x)$, where $\vec x \in \mathbb R^n$, and returns a vector:

$$\text{grad} \ f(\vec x) = \nabla f(\vec x) \equiv \sum_{i=1}^n \frac{\partial f (\vec x)}{\partial x_i} \vec e_i $$

where $\{\vec e_i \dots\vec e_n\}$ is an orthogonal basis of $\mathbb R^n$.

The divergence operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^n$, and returns a scalar function:

$$\text{div} \ \vec F(\vec x) = \nabla \cdot \vec F(\vec x) \equiv \sum_{i=1}^n \frac{\partial F_i (\vec x)}{\partial x_i} $$

The curl operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^3$, and returns a vector field:

$$\text{curl} \ \vec F(\vec x) = \nabla \times \vec F(\vec x) = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat i+\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat j+ \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat k$$

where $\hat i, \hat j, \hat k$ are the unit vectors of the three Cartesian axes.

Notice that, unlike the gradient and divergence, the curl operator does not generalize simply in $n$ dimensions. Also, the notation $\nabla \times \vec F$ is only a mnemonic device useful when we work in cartesian coordinates: in other coordinate systems, applying $\nabla \times \vec F$ will hold the wrong result.

We should probably also mention the laplacian operator, which is the divergence of the gradient:

$$\nabla^2 f(\vec x) \equiv \text{div} \ (\text{grad} \ f(\vec x)) = \nabla \cdot (\nabla f(\vec x))$$

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So, to sum up, $\nabla$ is just a useful notation that is used in the representation of three different vector operators. It turns out that we can often formally manipulate $\nabla$ as if it was a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless.

To see this, just consider one of the fundamental properties of vector spaces: if $v,w$ are elements of the vector space $V$, then $v+w$ is also an element of $V$.

Let's consider the vector space $\mathbb R^n$: what meaning should we give to an expression such as

$$\nabla + \vec x \ ?$$

the answer is: no meaning at all, because $\nabla$ is not a vector.

Both. It's an operator that transforms as a covector under rotations. What this means is that if you rotate the coordinate system the gradient in the new coordinate system, $\nabla'$, can be written as:$$\nabla'_i = \sum_{j} R^{-1}_{ij} \nabla_j,$$ where $R^{-1}$ is the inverse of the rotation matrix, $\nabla$ is the gradient in the original coordinate system, and $\nabla'$ is the gradient in the rotated coordinate system.