Let $N\mathcal C$ be the nerve of category. It's true that for $0<k<n$, every horn $\Lambda^n_k\to N\mathcal C$ can be extended to $\Delta^n$ in a unique way.
What about $k = 0$ or $n$? For $n \leq 2$, this certainly can't always be done (if $n=1$ this would imply $\mathcal C$ is the terminal category, and if $n=2$ that $\mathcal C$ is a groupoid).
However, for $n=3$, it seems that the horns can be filled even when $k = 0$ or $3$! For instance, a horn $\Lambda^3_0\to N\mathcal C$ would correspond to three triangles $\sigma_1,\sigma_2,\sigma_3\in N\mathcal C_2$ fitting in the picture below:
The tetrahedron is already filled with $\sigma_0 := (d_0\sigma_3,d_0\sigma_1,d_0\sigma_2)$ as the $0$-th face.
Am I making any mistake? Does this pattern goes on for $n\geq 4$ i.e. in this case can every horn be filled?
You can fill in the three boundary edges of the triangle $\sigma_0$. However, to fill in $\sigma_0$ itself (as well as the interior of the tetrahedron), you need to know that these three edges form a commutative diagram in your category. This is not necessarily true. For instance, if the $0$th object of the diagram is the initial object of your category, then you could put any three maps at all on the edges of $\sigma_0$, and the other three triangles will automatically commute, so you would be asserting that every triangular diagram in your category commutes!