Are functor categories with triangulated codomains themselves triangulated?
The statement is false.
For example, take $C=[1]\times [1]$ to be a square and $\mathcal{T} = h\mathsf{Sp}$ to be the homotopy category of spectra. Now consider the square $X$ with $X(0,0) = S^2$, $X(1,0) = S^1$, and the other values zero, and the other square $Y$ with $Y(1,0) = S^1$ and $Y(1,1) = S^0$. Take the maps $S^2 \to S^1$ and $S^1 \to S^0$ to be $\eta$, and consider the natural transformation $X \to Y$ which is given by multiplication by 2 on $X(1,0)=S^1 \to S^1 = Y(1,0)$.
If this map had a cofiber, then, from the initial to final vertex we would get a map $S^3 \to S^0$. Following the square one direction, we see that we would have some representative for the Toda bracket $\langle \eta, 2, \eta\rangle$. Following the other direction, we factor through zero. But this Toda bracket consists of the classes $2\nu$ and $-2\nu$; in particular, it does not contain zero.
[Of course, this example can be generalized to any nontrivial Toda bracket/Massey product in any triangulated category you're more familiar with.]
Indeed, the Toda bracket is exactly the obstruction to 'filling in the cube' for the natural transformation $X \to Y$.
Anyway- this is one of many reasons to drop triangulated categories in favor of one of the many modern alternatives (e.g. stable $\infty$-categories, derivators, etc.).
As for t-structures and so on, in the land of stable $\infty$-categories these are easy to come by. (See, e.g., Higher Algebra section 1.2.1 and Proposition 1.4.4.11 for various tricks for building these.)
Dylan Wilson's example is excellent. Let me offer another one, with a more algebraic and "finitistic" flavor.
In my opinion, the simplest triangulated category $\mathcal{T}$ is the category of finite-dimensional vector spaces over a field $k$, with identity suspension (a.k.a. translation) functor and $3$-periodic long exact sequences as exact triangles. (This is actually the only triangulated structure carried by $\mathcal{T}$ up to equivalence.)
Let $C_2$ be the cyclic group of order $2$ (regarded as a category with just one object). Then the functor category $\mathcal{T}^{C_2}$ is the category of finitely generated modules over the group algebra $k[C_2]$. This is the same as the category of finitely generated projective modules over the so-called Auslander algebra $B$ of $k[C_2]$. By a result Freyd, if $\mathcal{T}^{C_2}$ were triangulated then $B$ would be self-injective.
If $k$ has characteristic $2$, $k[C_2]\cong k[\epsilon]/(\epsilon^2)$ is the algebra of dual numbers and $B$ is the endomorphism algebra of the $k[\epsilon]/(\epsilon^2)$-module $k\oplus k[\epsilon]/(\epsilon^2)$. This $B$ is not self-injective. Indeed, since $k$ has characteristic $2$, $k[\epsilon]/(\epsilon^2)$ is not semi-simple, so $B$ has global dimension $2$. If $B$ were self-injective it would have global dimension either $0$ or $\infty$.
I believe I have a simpler counterexample, which I learned from Paul Balmer's course on tensor-triangular geometry last spring:
Claim The arrow category $\mathcal{T}^{\bullet \to \bullet}$ of a triangulated category $\mathcal{T}$ never has any triangulated structure unless $\mathcal{T} = 0$. Actually, we don't even need $\mathcal{T}$ to be triangulated here: if $\mathcal{T}$ is any additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T} = 0$.
Proof: Suppose $\mathcal{T}$ is an additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated. Let $a$ be an arbitrary object in $\mathcal{T}$, with identity morphism $1_a : a \to a$. Let $t$ denote the unique morphism $a \to 0$. Then $\require{AMScd}$ \begin{CD} a @>1_a>> a\\ @V 1_a V V @VV t V\\ a @>>t> 0 \end{CD} defines a morphism $\alpha : 1_a \to t$ in $\mathcal{T}^{\bullet \to \bullet}$. Note that $\alpha$ is an epimorphism. In any triangulated category, all epimorphisms are split, so let $\beta : t \to 1_a$ be a splitting of $\alpha$ (that is, $\alpha \circ \beta$ is the identity morphism of $t$). Then $\beta$ is a commutative diagram \begin{CD} a @>t>> 0\\ @V f V V @VVs V\\ a @>>1_a> a \end{CD} such that $1_a \circ f = 1_a$ (and $t \circ s = 1_0$). From this and the commutativity of the diagram, we see that $1_a = 1_a \circ f = s \circ t$ factors through $0$. Thus, $a = 0$. Since $a$ was arbitrary, $\mathcal{T} = 0$.
Edit: Of course we could make the statement even weaker: we only really needed that $\mathcal{T}$ has a zero object. But if $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T}$ must be additive, because it embeds as an additive subcategory of $\mathcal{T}^{\bullet \to \bullet}$ via $a \mapsto 1_a$.