What is the set of triangular numbers mod $n$? My question is motivated by the following fact (for a couple of proofs see my answer at MSE 1926967; note that this fact also answers the $\ell = 1$ case of MO 265513):

Fact. The set of triangular numbers modulo $n$ yields $\mathbb{Z}/n\mathbb{Z}$ iff $n$ is a power of $2$.

My specific question is whether there is a straightforward way to describe this set given, for example, the prime factorization of $n$. But I also have the broader question:

What else can be said about the set of triangular numbers modulo $n$ as a function of $n$?

For example, when $n$ is odd, we might expect the set to have cardinality around $(n+1)/2$. To illustrate this by way of example, consider $n = 11$, and the first $11$ counting numbers each reduced modulo $11$:
$$1, 2, 3, 4, 5, -5, -4, -3, -2, -1, 0$$
However many distinct elements we have after the first $(n+1)/2$ triangular numbers will be the final total, since (as suggested by the negative notation above) the elements reverse course thereafter, and wend their way back to $0$ at which point they reset as they have formed a cycle.
But I am not only interested in an estimate of the cardinality based on $n$'s parity; I am hoping for the whole shebang! In that spirit:

Q. How can we describe the set of triangular numbers modulo $n$ given $n$'s prime factorization?

 A: Fix $n=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$. We want to know for which $b$ the equation
$$
  a(a+1)/2=b
$$
has a solution modulo $n$. By the Chinese remainder theorem, this happens iff it has a solution mod $p_i^{\alpha_i}$ for each $i$. Consider one of these prime powers. If $p_i=2$ then there is always a solution, as mentioned in your question. Suppose $p_i\neq2$. The equation is equivalent to
$$
  (2a+1)^2=8b+1.
$$
Let $8b+1=x_ip_i^{\beta_i}$ where $p_i\not\mid x_i$. If $\beta_i\geq\alpha_i$ we can take $2a+1=p^{\alpha_i}$. Otherwise $\beta_i$ must be even, and in this case by Hensel's lemma, there is a solution iff $x_i$ is a quadratic residue mod $p_i$. Therefore the residues of triangular numbers mod $n$ are exactly
$$
  \left\{b\,\left|\,\left(\frac{x_i}{p_i}\right)=1\text{ and }\beta_i\text{ is even for each }i\text{ with }\beta_i<\alpha_i\text{ and }p_i\neq2\right.\right\}
$$
where $\left(\frac{x}{p}\right)$ is the Legendre symbol. I don't think there is a closed form expression for the Legendre symbol, but quadratic reciprocity gives an algorithm for calculating it.
