All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand. The result, though not phrased then in terms of Morse theory, has been known since the 30s. So in my post I meant "topological manifold" when I said "non-triangulable manifold", because that is the only thing that could have made sense! 

For dimensions $D \leq 4$ any triangulable manifold is smooth with no further requirement.  

For any dimension $D \geq 5$ there are spin manifolds which are triangulable, but do not admit any smooth structure. 

So the answer to your title question is $D \leq 4$, and no reasonable assumption on $M$ for $D \geq 5$ is likely to give you a similar result in higher dimensions. Certainly spin is not enough. 

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As for "triangulable iff smooth". The standard statement is that the notion of PL and smooth structures are equivalent in dimensions up to 6 (in dimension 7, every PL structure may be smoothed, but not uniquely). See [here](https://mathoverflow.net/q/96670/40804) for a good discussion. 

A PL manifold can be thought of as a topological manifold $X$ equipped with a(n equivalence class of) triangulation so that the [link](https://en.wikipedia.org/wiki/Triangulation_(topology)#Piecewise_linear_structures) of any vertex is PL-homeomorphic to $S^{\dim X - 1}$. 

In dimension 4 it is already difficult to prove that "triangulable and PL-able" are equivalent. In any dimension, it is not hard to show that $\text{lk}(\sigma)$ is a [homology manifold](https://en.wikipedia.org/wiki/Homology_manifold) using the cone-homeomorphism from $\text{lk}(\sigma) \times \Bbb R$ to an open subset of a topological manifold; in particular, $\text{lk}(\sigma) \times \Bbb R$ is a topological manifold. It is similarly not hard to show that $\text{lk}(\sigma)$ must have the same homotopy type as $S^{\dim M - 1}$. In dimensions $\leq 2$, homology manifolds are in fact manifolds (Theorem 16.32 in Bredon's sheaf theory) and topological manifolds admit unique PL structures, so we see that there is a PL homeomorphism $\text{lk}(\sigma) \cong S^{\dim M - 1}$. In particular, any triangulation of a manifold of dimension at most $3$ equipped is automatically PL.

In dimension 4 the same is true but one needs to do more work: now you need to use the fact that $\text{lk}(\sigma)$ is itself a simplicial complex. A theorem of Edwards (theorem 3.5 [here](https://www.maths.ed.ac.uk/~v1ranick/papers/cannon2.pdf)) then indicates that $\text{lk}(\sigma)$ is a manifold, and hence by the Poincare conjecture that it is homeomorphic to $S^3$. 

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What you want now is an example, in each dimension $n \geq 5$, of a triangulable manifold $X_n$ which is not PL (and hence not smooth). This is discussed on MO [here](https://mathoverflow.net/a/214447/40804), citing [Rudyak](https://arxiv.org/pdf/math/0105047.pdf): if $X_4$ is Freedman's E8 manifold, then $X_{4+k} = X_4 \times T^k$ is triangulable, but not PL. This is proved by a sort of dimensional reduction for both parts: in Rudyak's Theorem 7.4, he argues that none possess a PL structure by passing to the universal cover $\widetilde{X}_{4+k} = X_4 \times \Bbb R^k$. The Kirby-Siebenmann product theorem (relating PL structure sets on $M$ and $M \times \Bbb R$) states that this carries a PL structure for any $k \geq 1$ if and only if $\widetilde X_5$ does. Because PL 5-manifolds carry smooth structures, $\widetilde X_5$ is smoothable; one then argues by constructing a bordism between $X_4$ and a smooth spin manifold with the same signature, which is impossible by Rokhlin's theorem. Therefore no $\widetilde X_n$ is PL for $n \geq 4$, and hence neither is any $X_n$ for $n \geq 4$. 

To see that $X_n$ is triangulable for $n \geq 5$, see Rudyak's Theorem 21.5: that every compact orientable 5-manifold is triangulable is a theorem of Siebenmann. Then $X_{5 + k} = X_5 \times T^k$ is also triangulable, being a product of triangulable manifolds.

Here is a reasonable approach one might try to see that any orientable 5-manifold is triangulable, but I think it is circular, or at least circuitous.

That any orientable 5-manifold is triangulable might also follow (but see below) from [Galewski-Stern](https://www.maths.ed.ac.uk/~v1ranick/papers/galester.pdf). One of the relevant theorems is that if $$0 \to \text{ker}(\mu) \to \Theta \xrightarrow{\mu} \Bbb Z/2 \to 0$$ is the short exact sequence with $\Theta$ the 3-dimensional homology cobordism group and $\mu$ the Rokhlin homomorphism (take a spin manifold $X$ that a homology 3-sphere $\Sigma$ bounds; then $\mu([\Sigma]) = \sigma(X)/8 \mod 2$), then if $\Delta(M) \in \Bbb H^4(M; \Bbb Z/2)$ is the Kirby-Siebenmann class and $\beta_\Theta: H^*(M;\Bbb Z/2) \to H^{*+1}(M; \text{ker}(\mu))$ the associated Bockstein map, then a closed manifold $M$ of dimension $n \geq 5$ is triangulable if and only if $\beta_\Theta \Delta(M) = 0 \in H^5(M;\text{ker}(\mu))$. 

(Manolescu's [2013 contribution] was essentially that $\beta_\Theta$ is not identically zero, which is equivalent to the group theoretic statement that $\mu: \Theta \to \Bbb Z/2$ has no section.)

Any short exact sequence $0 \to H \to G \to K \to 0$ gives rise to a long exact sequence on cohomology (with boundary map the Bockstein). If $M$ is an oriented closed manifold of dimension $n$, then the end of this sequence is precisely $$H^{n-1}(M;K) \xrightarrow{\beta} H^n(M; H) \to H^n(M; G) \to H^n(M; K) \to 0;$$ because $H^n(M; A) \cong A$ naturally for an oriented closed $n$-manifold, the end of this sequence is precisely our original short exact sequence $H \to G \to K \to 0$; by exactness, we see that $\beta: H^{n-1}(M; K) \to H^n(M; H)$ is identically zero. 

In particular, if $M$ is a closed oriented 5-manifold, we must have $\beta_\Theta \Delta(M) = 0 \in H^5(M; \text{ker} (\mu))$. Therefore, $M$ is triangulable. 

But... the Galewski-Stern paper relies on the Siebenmann paper in which compact oriented 5-manifolds are shown more quickly to be triangulable.