Let me interpret $\mathcal{M}_{0,n}$ as the moduli space of genus zero curves with $n$ marked points, as opposed to punctures. This is just a psychological thing and should not make a difference to what follows.

That $\dim(\mathcal{M}_{0,n}) = 0$ for $n = 0,1,2,3$ is very classical: this amounts to saying that all genus zero curves (say over an algebraically closed field) are isomorphic to $\mathbf{P}^1$, and that given two pairs $(p_i)$ and $(q_i)$ of ordered tuples consisting of at most $3$ points, there is an automorphism of $\mathbf{P}^1$ sending the $(p_i)$ to the $(q_i)$. This automorphism can be constructed directly once you choose coordinates on $\mathbf{P}^1$.

Okay, now suppose that $n \geq 4$. A point of $\mathcal{M}_{0,n}$ is to be thought of as $\mathbf{P}^1$ together with $n$-points $p_1,\ldots,p_n$, all of this up to automorphisms of $\mathbf{P}^1$. As discussed in the previous paragraph, there is a unique automorphism of $\mathbf{P}^1$ that sends the first three points $p_1,p_2,p_3$ to $0,1,\infty$, in that order. So, spending this piece of freedom, a point of $\mathcal{M}_{0,n}$ can always be represented by $\mathbf{P}^1$ together with the $n$-points $0,1,\infty,p_4,\ldots,p_n$, where now $p_4,\ldots,p_n$ are some $n - 3$ distinct points of $\mathbf{P}^1$ different from $0,1,\infty$. In fact, by uniqueness of the automorphism sending $p_1,p_2,p_3$ to $0,1,\infty$, we see that every point of $\mathcal{M}_{0,n}$ has a *unique* representative of the form
$$
(\mathbf{P}^1, 0,1,\infty,p_4,\ldots,p_n)
\quad
p_4,\ldots,p_n \in \mathbf{P}^1 \setminus \{0,1,\infty\}\;\text{distinct.}
$$
So now we can compute dimensions by figuring out what freedom we have left: the only things that can move are the $n - 3$ points $p_4,\ldots,p_n$. Each $p_i$ is allowed to take all but finitely many values in $\mathbf{P}^1$. Since removing any finite number of points in $\mathbf{P}^1$ still gives a $1$-dimensional object, we see that $\dim(\mathcal{M}_{0,n}) = n - 3$, as you computed.