This is an expansion of Jason's comment.
The category of short exact sequences is obviously additive and has arbitrary coproducts. Thus, cocompleteness is equivalent to the existence of cokernels. So let us given a morphism $f_* : A_* \to B_*$ of short exact sequences:
$\begin{array}{ccccccccc} 0 & \rightarrow & A_1 & \rightarrow & A_2 & \rightarrow & A_3 & \rightarrow & 0 \\ & & ~ ~ \downarrow f_1 & &~~ \downarrow f_2 & & ~~ \downarrow f_3 & & \\ 0 & \rightarrow & B_1 & \rightarrow & B_2 & \rightarrow & B_3 & \rightarrow & 0 \end{array}$
The snake lemma gives us an exact sequence
$0 \to \mathrm{ker}(f_1) \to \mathrm{ker}(f_2) \to \mathrm{ker}(f_3) \to \mathrm{coker}(f_1) \to \mathrm{coker}(f_2) \to \mathrm{coker}(f_3) \to 0.$
Let $K$ be the kernel of $\mathrm{coker}(f_1) \to \mathrm{coker}(f_2)$. Equivalently, $K \cong \mathrm{ker}(f_3) / (\mathrm{ker}(f_2) / \mathrm{ker}(f_1))$. Then we have a short exact sequence $C_*$ defined by
$0 \to \mathrm{coker}(f_1) /K \to \mathrm{coker}(f_2) \to \mathrm{coker}(f_3) \to 0$
together with a morphism $p_* : B_* \to C_*$. I claim that this is the cokernel of $f_*$. It is an epimorphism since it components are epimorphisms. It factors through the non-exact cokernel of $f_*$, therefore we have $p_* f_* = 0$. Now let $D_*$ be another exact sequence and $g_* : B_* \to D_*$ be a morphism satisfying $g_* f_*$. Then $g_*$ factors through the non-exact cokernel of $f_*$.
We are left to prove that $g_1 : B_1 \to D_1$ vanishes on $K$, so that it even factors through $C_1= \mathrm{coker}(f_1) /K$. But this is a diagram chase: An element in $K$ is represented by an element in $B_1$ whose image in $B_2$ comes from an element in $A_2$. Thus the image in $D_2$ vanishes. Since $0 \to D_1 \to D_2$ is exact, this means that the image in $D_1$ already vanishes, qed.
Of course this reasoning can also be done with arrows. Therefore, if $\mathcal{A}$ is an arbitrary abelian category, then the category $S(\mathcal{A})$ of short exact sequences in $\mathcal{A}$ has cokernels and (by duality also) kernels. If $\mathcal{A}$ has coproducts and satisfies has AB4, then $S(\mathcal{A})$ has arbitrary coproducts and is therefore cocomplete.
The category of short exact sequences is never abelian, in fact not balanced: From the above description of cokernels we see that $f_*$ is an epimorphism in that category iff $f_2,f_3$ are epimorphisms. Intuitively it is ok that $f_1$ doesn't appear here since $f_1$ is uniquely determined by $f_2,f_3$ due to the exactness! Similarily, $f_*$ is a monomorphism iff $f_1,f_2$ are monomorphisms. Thus, $f_*$ is a mono- and an epimorphism iff $f_2$ is an isomorphism, $f_1$ is a monomorphism and $f_3$ is an epimorphism. However, $f_*$ is an isomorphism iff $f_1,f_2,f_3$ are isomorphisms.
Addendum.
We can simplify these arguments a lot: Consider the category $E(\mathcal{A}) \subseteq \mathrm{Mor}(\mathcal{A})$ of epimorphisms in $\mathcal{A}$. The morphisms are commutative diagrams. Obviously, coproducts and cokernels exist in this category, since the category is closed under these operations in $\mathrm{Mor}(\mathcal{A})$. There is a forgetful functor $S(\mathcal{A}) \to E(\mathcal{A})$, which turns out to be an equivalence of categories! The quasi-inverse chooses for every epimorphism $A_2 \to A_3 \to 0$ a kernel $0 \to A_1 \to A_2$. For every morphism $(f_2,f_3) : A_* \to B_*$ in $E(\mathcal{A})$ the universal property of the kernel yields a unique $f_1 : A_1 \to B_1$ such that $(f_1,f_2,f_3)$ becomes a morphism in $S(\mathcal{A})$. Since $E(\mathcal{A})$ is cocomplete, the same is true for $S(\mathcal{A})$. If one unwinds the definitions, one gets the same cokernels as described above explicitly.
Yes, right derived functors $R^nTA$ come automatically with two long exact sequences, one from varying $A$ and one from varying the functor $T$. This is discussed in chapter IV.6. of Hilton and Stammbach's book A Course in Homological Algebra. What follows is essentially me summarizing the content of this chapter.
Your notion of exactness of a sequence of functors is not the most efficient one. Let's say a sequence of natural transformations $T'\to T\to T''$ is short exact at injectives if and only if \begin{align*}0\to T'I \to TI\to T''I\to 0\end{align*} is exact for every injective object $I$. Given a sequence of functors as above which is short exact at injectives, fix an injective resolution $A\to \mathbf I$ of $A$. The natural transformations yield a sequence of chain maps
\begin{align*} 0\to T'\mathbf I \to T\mathbf I \to T''\mathbf I \to 0\end{align*}
which is by assumption levelwise short exact and thus short exact. Hence we get a long exact sequence in cohomology.
\begin{align*} 0 \to R^0T'A \to R^0TA \to R^0T''A\xrightarrow{\omega_0} R^1T'A \to R^1TA\to R^1T''A\to...\end{align*}
The long exact sequence is natural in $T$ and in $A$ in the sense desecribed in Hilton & Stammbach IV.6. Proposition 6.4. You do not need to know that the functors $T',T$ and $T''$ are left exact to get the long exact sequence. Your example does work. The most important examples are actually the classical derived functors $\mathrm{Tor}$ and $\mathrm {Ext}$.
Example. Let $\mathcal A$ be any abelian category with enough injectives. Define $\mathrm{Ext}_{\mathcal A}^n(A,-) $ as the right derived functor of $\mathrm{Hom}_{\mathcal A}(A,-)$. In other words
\begin{align*} \mathrm{Ext}_{\mathcal A}^n(A,B) = R^n\mathrm{Hom}_{\mathcal A}(A,-)B\,.\end{align*}
Given a short exact sequence $0\to B'\to B\to B''\to 0$ you obtain a long exact sequence in the second variable by the standard argument involving the Horseshoe lemma. But we can obtain also a long exact sequence in the second variable by using the method described above. A morphism $A'\to A$ yields a natural transformation $\mathrm{Hom}_{\mathcal A}(A',-)\to \mathrm{Hom}_{\mathcal A}(A,-)$. Thus a short exact sequence $0\to A'\to A\to A''\to 0$ yieds a sequence of natural transformations
\begin{align*}
\mathrm{Hom}_{\mathcal A}(A'',-) \to \mathrm{Hom}_{\mathcal A}(A,-)\to \mathrm{Hom}_{\mathcal A}(A',-)
\end{align*}
This sequence is short exact at injectives by definition of injective objects (but it is not short exact in your sense!). Applying the discussion above to the sequence yields the long exact sequence of $\mathrm{Ext}_{\mathcal A}^n$ in the first variable. In a similar way one can show that $\mathrm{Tor}_n^\Lambda(A,B) = L_n(A\otimes_\Lambda -)B$ has two kinds of long exact sequences. One is obtained by varying $A$ and one by varying $B$.
Additionally, since a morphism $A\to A'$ induces a natural transformation $\textrm{Hom}_{\mathcal A}(A',-) \to \textrm{Hom}_{\mathcal A}(A,-)$ and thus a map $\mathrm{Ext}_{\mathcal A}^n(A',B) \to \mathrm{Ext}_{\mathcal A}^n(A,B)$ we have found an easy way to show that $\mathrm{Ext}_{\mathcal A}^n$ is a bifunctor, without having to derive $\mathrm{Hom}_{\mathcal A}$ in the first variable and having to show that $\mathrm{Ext}_{\mathcal A}^n$ is balanced first. I have seen people doing it this way, and it is a mess. Both of the two long exact sequences of $\mathrm{Ext}_{\mathcal A}^n$ vary naturally in both $A$ and $B$.
Best Answer
This analogy can be made precise using the notion of homotopy limits, specifically the notion of homotopy fiber and fiber sequence. In short, in any higher category with zero objects (e.g. either chain complexes or pointed spaces), given a morphism $f : E \to B$, repeatedly taking homotopy fibers (the homotopy pullback of the diagram $E \rightarrow B \leftarrow \bullet$, where $\bullet$ is the zero object) gives a "long fiber sequence"
$$\cdots \to \Omega F \to \Omega E \to \Omega B \to F \to E \to B$$
where every object is the homotopy fiber of the preceding morphism. This construction, after possibly applying some auxiliary functors, is responsible for all long exact sequences in mathematics. There is a dual construction involving taking homotopy cofibers (which are certain homotopy pushouts), but it's just this construction in the opposite category, although it's sometimes done without zero objects (you can do it in spaces, not pointed spaces, using the point, which is the terminal object). That looks like
$$A \to B \to C \to \Sigma A \to \Sigma B \to \Sigma C \to \cdots$$
and it's the ability to form these "long cofiber sequences" that triangulated categories imperfectly attempt to capture.
In particular, this curious period-$3$ behavior in both of the above sequences is not an artifact and hasn't been put in by hand; it falls naturally out of higher-categorical universal properties.
What's special about short exact sequences with three terms is that the first term is the homotopy fiber of the morphism between the second two terms, or equivalently (this is special to "stable" contexts like chain complexes and spectra) the third term is the homotopy cofiber of the morphism between the first two terms.
An exact sequence with more than three terms gives rise to a spectral sequence. (Actually this is true even without exactness, although I think exactness makes the spectral sequence nicer.)