This answer is based on $\phi = \phi_0 \cos(\omega_0 t)$ being true.
EXPRESSION OF VELOCITY BY KINEMATICS
It is important to note that $l\dot\phi \neq l\omega_0 \phi$. You find this to be the case whenever you find $\dot\phi$ by differentiation.
$$\dot\phi = \frac{d}{dt}(\phi)=\frac{d}{dt}(\phi_0\cos(\omega_0t))$$
$$= -\phi_0 \omega_0 \sin(\omega_0 t)$$
$$ \therefore v = -l\phi_0 \omega_0 \sin(\omega_0 t)$$
The $\sin(\omega_0 t)$ is slightly problematic when we want to express $v$ as a function of $\phi$ because we still have a function of $t$. Let's find a work around:
$$v^2 = l^2 \phi_0^2 \omega_0^2 \sin^2(\omega_0 t)$$
$$= l^2 \phi_0^2 \omega_0^2 (1-\cos^2(\omega_0 t))$$
$$= l^2 \omega_0^2 (\phi_0^2 - \phi_0^2 \cos^2(\omega_0 t))$$
$$= l^2 \omega_0^2(\phi_0^2 - \phi^2)$$
Therefore, the magnitude of the velocity is given by:
$$|v| = l \omega_0 \sqrt{\phi_0^2 - \phi^2}$$
EXPRESSION OF VELOCITY BY CONSERVATION OF ENERGY
However, as Phoenix87 points out, this is only an approximation for small $\phi$. This means that the equation for the conservation of energy will only be equal if you also approximate the conservation of energy equation:
$$1 - cos(x) \approx \frac{1}{2} x^2 $$
$$\therefore mgl(\frac{1}{2} \phi_0^2) = mgl(\frac{1}{2} \phi^2) + \frac{1}{2} mv^2$$
$$v^2 = gl(\phi_0^2 - \phi^2)$$
$$|v| = \sqrt{gl}\sqrt{\phi_0^2 - \phi^2}$$
Note that from first principles (see below), that, by approximation, $\omega_0^2 = g/l$, so:
$$|v| = l\omega_0 \sqrt{\phi_0^2 - \phi^2}$$
So we get the same expressions for the magnitude of $v$.
DERIVATION OF APPROXIMATE EXPRESSION FOR PHI
To see where the approximation of $\phi = \phi_0 \cos(\omega_0 t)$ comes, we could look at the derivation of the differential equation for $\phi$:
The torque, $\tau$, about the point of attachment of the pendulum to the ceiling is given by:
$$\tau = -mgl \sin(\phi)$$
The rotational variation of Newton's 2nd Law is:
$$\tau = J \ddot \phi $$
where $J$ is the moment of inertia about the point of attachment of the pendulum (analogous to the mass in $F = ma$).
$$J = ml^2$$
$$\therefore -mgl\sin(\phi) = ml^2\ddot \phi$$
So, we get:
$$l\ddot \phi + g \sin(\phi) = 0$$
This is not a nice equation to solve at all, so... we linearise! We assume $\phi$ to be small, so $\sin(\phi) \approx \phi$.
So,
$$l\ddot \phi + g \phi = 0$$
And, if you solve this the way you solve other linear second order differential equations, you get:
$$\phi = A\sin\left(\sqrt{\frac{g}{l}}t\right) + B\cos\left(\sqrt{\frac{g}{l}}t\right)$$
where $A$ and $B$ are arbitrary constants that we will determine by two boundary conditions, where the pendulum is initially held at rest at angle $\phi_0$:
$$t = 0, \phi = \phi_0$$
and
$$t = 0, \dot \phi = 0$$
When we substitute the first boundary condition, we get:
$$\phi_0 = B$$
To make use of the second boundary condition, we differentiate the expression for $\phi$ wrt to $t$ to get:
$$\dot\phi = A\sqrt{\frac{g}{l}}\cos\left(\sqrt{\frac{g}{l}}t\right) - B\sqrt{\frac{g}{l}}\sin\left(\sqrt{\frac{g}{l}}t\right)$$
So, substituting the second boundary condition gets us:
$$0 = A\sqrt{\frac{g}{l}}$$
$$\therefore A = 0$$
We have now determine values for the two arbitrary constants, so we finally get the approximate expression for $\phi$ that you started off with:
$$\phi = \phi_0 \cos\left(\sqrt{\frac{g}{l}} t\right)$$
For the sake of convenience, we let $\sqrt{\frac{g}{l}} = \omega_0$ (hence the use of $\omega_0^2 = g/l$ above).
$$\phi = \phi_0 \cos(\omega_0 t)$$
