The magnetic field along the axis of a long solenoid comes out to be $\mu_0nI$ using ampere's circuital law, considering a rectangular amperian loop.
But if I try to apply the law to an equivalent bar magnet using the same loop, there is no electric current through the loop, but $\int \overrightarrow B. \overrightarrow {dl}$ is non zero for $\overrightarrow B$ along the magnet's axis. What am I missing? Is the law even applicable here?
The solenoid and the bar magnet generate their magnetic fields using different mechanisms, so the solenoid equation cannot be applied to a bar magnet.
A moving charge, like an electron, generates a magnetic field, and a current is just a load of charges moving in a conductor. That's why a current generates a magnetic field. The equation for the solenoid is derived using Ampère's law for the specific geometry of a solenoid and feeding in the current in the wire.
However even a stationary electron has a (tiny) magnetic field due to its spin. This field is always present whether or not the electron is moving. This is called its magnetic moment. In most cases the electrons have their fields arranged randomly so their fields cancel out and the total magnetic field is zero. However in a class of materials known as ferromagnets the electron fields line up with each other to generate a large total field, and that's what happens in a bar magnet. Since this field has nothing to do with currents you cannot use Ampère's law to calculate it.
The Ampere's law applies to magnetic H field, not magnetic B field because $\nabla\times\vec{H}=\vec{J}$, where $\vec{J}$ is conducting current. The $\mu_0nI$ in the first paragraph of your statement should be changed to $nI$. If there is only a magnet and the conduction current is 0, then $\nabla\times\vec{H}=0$, so $\oint\vec{H}\cdot d\vec{l}=0$, which is consistent with the Ampere's law.
For a bar magnet, there is a 'bound surface current' on its surface given by $\bf K_{\rm bound}=M\times{\hat n}$. This acts like a current in a solenoid.
