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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.

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

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.

In dimensions at least 5 you should expect this to be false, because if $P$ is a homology sphere of dimension $n \geq 3$, then $\Sigma^2 P \cong S^{\dim P + 2}$ is naturally triangulated (you may always suspend a triangulation), but it is not a PL structure: the links of the top/bottom vertices are $\Sigma P$, which is not a manifold. Of course, the sphere is PL-able, so this is not a counterexample to your question. But it indicates there's no reason to believe there isn't one - the links of vertices are not automatically spheres for silly reasons anymore.

Now apply Galewski-Stern, which says the following (though note the discussion of the Kirby-Siebenmann invariant is more classical than their paper!).

There is a group $\Theta$, called the homology cobordism group, which fits into a short exact sequence $0 \to \text{ker}(\mu) \to \Theta \to \Bbb Z/2 \to 0$. There is an invariant $\text{ks}(M) \in H^4(M;\Bbb Z/2)$ of topological $n$-manifolds, and if $n \geq 5$ the vanishing $\text{ks}(M) = 0$ is equivalent to $M$ supporting a PL structure. Assuming still that $n \geq 5$ and writing $\beta_{\Theta}: H^4(M;\Bbb Z/2) \to H^5(M;\text{ker}(\mu))$ for the associated Bockstein map, then $\beta_{\Theta} \text{ks}(M)$ vanishes if and only $M$ is triangulable.

The fact that $\beta_\Theta$ is nonzero is essentially Manolescu's 2012 contribution to the triangulation question.

So what you want to find is an example of a $(\geq 5)$-dimensional $M$ so that $\text{ks}(M) \neq 0$ but $\beta_\Theta \text{ks}(M) = 0$. Take $X$ to be the $E8$ manifold, and let $X_k = X \times T^{4 - k}$ be the $k$-dimensional manifold you get by crossing with enough circles. We already know that $\text{ks}(X) \neq 0$, because for any 4-manifold with even intersection form, $\text{ks}(X) = \sigma(X)/8 \pmod 2$, and $\sigma(X) = 8$ in this case. Because Kirby-Siebenmann invariants of products $X \times Y$ are (under the Kunneth decomposition) the same as $\text{ks}(X) + \text{ks}(Y)$, we see that $\text{ks}(X_k) = \text{ks}(X) \neq 0$ for all $k$. But because $\beta_\Theta H^*(X;\Bbb Z/2) = 0$, and $\beta_\Theta$ acts on $H^*(X \times Y;\Bbb Z/2) = H^*(X;\Bbb Z/2) \otimes H^*(Y;\Bbb Z/2)$ via $\beta_\Theta \otimes 1 + 1 \otimes \beta_\Theta$, we see that $\beta_\Theta \text{ks}(X_k) = \beta_\Theta \text{ks}(X) = 0$ for all $k \geq 4$.

Thus by the Galewski-Stern result (and Kirby-Siebenmann's work on PL structures), $X_k$ is triangulable for every $k \geq 5$ but not PL. What is more, because $w_1(X_k) = w_2(X_k)$, these are spinnable, if you so desire.

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, citing Rudyak: 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 orientable 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. 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.