The wave function isn't an operator; the word "operator" in quantum mechanics means something more precise than "function". You might say that the momentum operator is "something to put in an integral to get the expectation value of momentum"; let me explain.
We know that for a particle in a state $\psi$,
$$
\langle x \rangle = \int_{-\infty}^{+\infty} x |\psi(x,t)|^2 \mathrm{d}x
$$
because $|\psi(x,t)|^2\mathrm{d}x$ is the probability that the particle will be found in the small interval $(x,x+\mathrm{d}x)$ at time $t$. Let's differentiate this and push the derivative into the integral
$$
\frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \int_{-\infty}^{+\infty} x \frac{\partial|\psi(x,t)|^2}{\partial t} \mathrm{d}x
$$
Now we use Schrodinger's equation to replace the time derivative with space derivatives
$$
\frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \frac{i\hbar}{2m}\int x \frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x
$$
Use integration by parts to eliminate integrate away the outer $\frac{\partial}{\partial x}$ and differentiate the $x$
$$
\frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{2m}\int \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x
$$
Perform another integration by parts on the second term
$$
\frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{m}\int \left(\psi^*\frac{\partial\psi}{\partial x}\right) \mathrm{d}x
$$
Multiply by $m$
$$
\langle p \rangle = \int \psi^* \left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right) \psi \mathrm{d}x
$$
Notice that this is the same form as the integral for $\langle x \rangle$, except we use $\frac{\hbar}{i}\frac{\partial}{\partial x}$ in place of $x$. That's why they're called the momentum and position operators respectively; they are the operators you place between $\psi^*$ and $\psi$ in the integral to obtain the expectation value of that variable. There are also operators for angular momentum, energy, etc.
Of course, there are other more useful definitions of an operator. For example, notice that if we write $\psi$ as a sum of sinusoidal functions, each sinusoidal function is an eigenfunction of the momentum operator, and the integral becomes very simple to evaluate; hence we could think of the momentum operator as an operator that returns the sum of the component sinusoidal functions (momentum eigenstates) of $\psi$, weighted by momentum, which is how I like to think of it as.
Best Answer
The simplest possible solution to the Schrodinger equation is that of a plane wave: $$ \psi = e^{i(\boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{r}}- \omega t )} $$ In this scenario, the position of the system is indeterminate, but the momentum is known. Next, take the gradient of this wavefunction. $$ \nabla \psi = ik \psi$$ But we know that $p = \hbar k $, from De Broglie. Substituting this into the equation and rearranging yields $$ \frac{\hbar}{i} \nabla \psi = p \psi $$ This is an example of an eigenvalue equation - an operator acts on a function or vector to produce a constant multiple of the original function. So, we see that the eigenvalue is the measured momentum. Therefore, the operator producing it, $ \frac{\hbar}{i} \nabla $, is called the momentum operator.