Is $\nabla$ a vector? The following passage has been extracted from the book "Mathematical methods for Physicists":  

A key idea of the present chapter is that a quantity that is properly called a vector must have the transformation properties that preserve its essential features under coordinate transformation; there exist quantities with direction and magnitude that do not transform appropriately and hence are not vectors.  

Cross product: $\nabla \times (Vector)=Vector$   
From the above equation of cross product we can say that $\nabla$ is a vector (specifically vector operator). However, a vector generally has magnitude and an associated direction. While in case of $\nabla$, it might satisfy essential features under transformation to be a vector, but I don't see whether it has magnitude or not? Does it has magnitude? If so, what is it? Or otherwise is it that a vector need not have magnitude?  

With $f$ a scalar function of the coordinates, $\nabla f$ is a vector called the gradient of $f$.
With $f$ a vector function of the coordinates, $\nabla.f$ is a scalar called the divergence of $f$.
With $f$ a vector function of the coordinates, $\nabla\times f$ is a vector called the curl of $f$.
These three symbols ($\nabla,\nabla.,\nabla\times$) are differential operators and represent no quantity by themselves.
If you really want to see $\nabla$ as a vector, then it is $$\nabla=i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}$$
