Realizing universal $C^*$-algebras as concrete $C^*$-algebras

Edit: as pointed out in the comments, the following answers the question for unital C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra.

It may depend on what exactly you mean by "concrete", but I highly doubt that there is a general solution to this; finding a concrete realization of a universal C*-algebra requires classifying all the representations of the given generators and relations on Hilbert space, and this is an extremely difficult problem in general. For a more rigorous argument, see below.

But in your four examples, the universal C*-algebras are all commutative, and simple answers are possible:

1. $C([-1,1])$.
2. $C([0,1])$.
3. $C(\mathbb{D})$ where $\mathbb{D}$ is the unit disk.
4. $C(\{0,1\})=\mathbb{C}^2$.

In each case, the generator is the identity function, just as in your $C(\mathbb{T})$ example. It is a good exercise to verify the required universal property in each of these cases.

Another good example is the C*-algebra freely generated by two projections. This turns out to be the group C*-algebra $C^*(\mathbb{Z}_2\ast \mathbb{Z}_2)=C^*(\mathbb{Z}\rtimes\mathbb{Z}_2)$ and can be realized concretely as the subalgebra of $C([0,1],M_2(\mathbb{C}))$ containing those matrix-valued functions which are diagonal on the endpoints $0$ and $1$. See this paper of Raeburn and Sinclair.

So why do I think that a general solution is impossible? Consider the word problem for groups: there are groups given in terms of generators and relations for which there is no algorithm that can decide whether a given word in the generators represents the unit element. Now we can look at the maximal group C*-algebra of such a group. This C*-algebra is itself given by the same generators and relations together with additional relations requiring the generators to be unitary. If your intended meaning of a "concrete representation" comprises the existence of an algorithm that decides whether a given formal combination of generators represents $0$, then it follows that such a concrete representation cannot exist.

Just a supplement to the answer by Tobias Fritz: All your examples are obviously commutative, since there is only one generator which is normal. Thus the question is really about finding certain terminal compact Hausdorff spaces. For example 1. comes from the terminal compact Hausdorff space $X$ equipped with a continuous function $X \to \mathbb{C}$ which is self-adjoint and norm $1$, i.e. whose image equals $[-1,1]$. This is obviously $[-1,1]$, equipped with the identity. You get the same answer when the norm is supposed to be $\leq 1$ (but $<1$ doesn't work). In a similar way one gets the other answers mentioned by Tobias Fritz.

Here is a further supplement: a tip for how to check if a $C^\ast$-algebra $A$ is the universal $C^\ast$-algebra for a given presentation. (I probably learned this from Terry Loring's book "Lifting solutions to perturbing problems in $C^\ast$-algebras".)

First, check that $A$ really is generated by a set of elements satisfying the given relations.

Second, check that every irreducible representation of the universal $C^\ast$-algebra is a representation of $A$. Say your generators are $x_1,\dots,x_n$. Then an irreducible representation would be generated by elements $X_1,\dots,X_n$. Since the centre of an irreducible representation is trivial, anything built out of the $X_i$'s that $*$-commutes with all the $X_i$'s is a scalar - so this approach works well if your relations entail a certain amount of commutativity, since commuting elements.

For example, to show that the universal $C^*$-algebra on a self-adjoint element of norm at most $1$ is $C_0([-1,1] \setminus \{0\})$, the second part above would go as follows. Let $X$ be the generator in an irreducible representation. Then $X$ is a scalar, which is self-adjoint (i.e. real) and has norm at most $1$. So $X = t \in [-1,1]$, and this representation corresponds to evaluating at this point $t$.