Using the universal characterisation to derive properties of the quotient group and projection Let $N$ be a normal subgroup of some group $G$. If I understand it correctly a group $Q$ and a homomorphism $\pi \colon G \to Q$ are a quotient group of $G$ by $N$ if for any group $H$ and homomorphism $\phi \colon G \to H$ with $N \subseteq \ker(\phi)$ there is a a unique homomorphism $\psi \colon Q \to H$ with $\phi = \psi \circ \pi$.
I was wondering if this characterization is enough to prove that $\ker(\pi)=N$ and $\pi(G)=Q$ or if details of the standard quotient construction in terms of cosets is required to prove these facts.
 A: Mac Lane doesn't just claim this, he cites the book Algebra by Garret Birkhoff and Saunders Mac Lane (not to be confused with their earlier book A Survey of Modern Algebra) wherein the proofs are carried out (more precisely, pp. 80,81,410-412). The only aspect of normality used in the proofs (at least upon my skimming it) is that normal subgroups are exactly the kernels. In other words, if you replace "normal subgroup" with "kernel of a group homomorphism" in the various statements, the proofs should go through word for word. From this point of view, the coset construction is only a characterization of which subgroups are kernels.
For example, if $N$ is the kernel of $\phi\colon G\to L$, then the universal property of the quotient $G\to G/N$ implies that $N$ is in the kernel of $G/N$ and that we have a factorization $G\to G/N\to L$ of $G\to L$ for a unique $G/N\to L$. But then the kernel of $G\to L$, which is $N$, is contains the kernel of $G/N$, which contains $N$, so $N$ is the kernel of $G/N$.
A more categorical understanding of normality does exist. Namely, a monomorphism $N\hookrightarrow X$ is normal if it is contained in an equivalence class of an equivalence relation on $X$. For more details (though pitched at a somewhat advanced level), see the book From Groups to Categorical Algebra - Introduction to Protomodular and Mal'tsev Categories, by Dominuqe Bourn, or for an even more advanced text, the book Mal'cev, Protomodular, Homological and Semiabelian Categories by Francis Borceux and Dominique Bourn. However, unless the category is exact, normal subgroup are not necessarily kernel, e.g. for Hausdorff topological group kernels are closed normal subgroups.
A: Let $\mathcal{C}$ be a category with an object $0\in C$ that is both initial and terminal.
Then for any objects $X,Y$ we have a unique morphism $X\to Y$ which factors through $0$.  We call this morphism trivial.
Given a morphism $n\colon N \to G$, define its cokernel (if it exists) to be a morphism $\pi\colon G\to Q$ such that $\pi n$ is trivial, and given any morphism $\phi\colon G \to H$, with $\phi n$ trivial, we have a unique morphism $\psi\colon  Q\to H$ with $\psi\circ\pi=\phi$.
$$\begin{array}{cccc}
N & \stackrel n\to & G & \stackrel \pi \to & Q\\
&&& \stackrel \phi\searrow&\downarrow{\tiny \psi}\\
&&&&H
\end{array}$$
Your second question is can we deduce $\pi(G)=Q$.  What does this mean in a general category?  Certainly the codomain of $\pi$ is $Q$.  However you want to say that the image of $\pi$ is the whole of $Q$.  This is equivalent to saying that $\pi$ is a surjective group homomorphism, which is equivalent to saying that $\pi$ is epic.  This last statement is well defined in an arbitrary category, and moreover we can prove that it is true in this case.
Given $f,g\colon Q\to H$ such that $f\pi=g\pi$ we can set $\phi=f\pi=g\pi$.  Then $\phi n$ is trivial, so $\phi$ factors uniquely through $\pi$.  Thus $f=g$.
Now for your first question.  Given $n,\pi$ as before, we say that $n$ is the kernel of $\pi$ precisely when given any map $p\colon P\to G$ with $\pi p$ trivial, we have a unique map $w\colon P\to N$ such that $p=nw$.
$$\begin{array}{cccc}
N & \stackrel n\to & G & \stackrel \pi \to & Q\\
 \uparrow{\tiny w}&\nearrow{\tiny p}\\
P\end{array}$$
Now your first question is can we deduce that $n$ is the kernel of $\pi$, assuming $n$ is normal?
This brings us to the question of what do we mean by $n$ is normal?  A good starting place is that $n$ be monic.  We could try adding that its cokernel exists.  However all of these are true for the inclusion of an arbitrary subgroup in a group.  The cokernel is then the quotient of the normal closure of the subgroup.
What we can do is define $n$ to be normal precisely when its cokernel exists, and $n$ is the kernel of this cokernel.  This works - the normal morphisms in the category of groups are then precisely the inclusions of normal subgroups.
Your first question now becomes rather easy.  We have that the kernel of $\pi$ is $n$, by definition of $n$ being normal.
In summary, in any category with $0$ object, the relevant notions can be defined in a way that precisely agrees with the standard definitions in the category of groups, and with respect to these definitions your two statements can be deduced.

An alternative definition of normal is to say that $n$ is normal if it is the kernel of some map.  This also agrees with the usual definition of normal subgroup, in the category of groups.  Proving that $n$ is the kernel of $\pi$ is  still trivial (see comment below by @VladimirSotirov).
