It's a well-known theorem by Beilinson [1] that states that for each $n$ there exists a finite dimensional-algebra $B_n$ such that there is an equivalence of triangulated categories $D^\text{b}(\text{coh}\ \mathbb{P}^n)\cong D^\text{b}(\text{mod}\ B_n)$.
$B_1$ is the path algebra of the Kronecker quiver, and hence has global dimension 1. In general $B_n$ are quiver algebras. I've tried to calculate for the cases $n=2$ and $n=3$, and have found out that these $B_n$'s are also of finite global dimension. So it's natural to ask whether all $B_n$'s have finite global dimensions.
Is there a general proposition that sounds like:
Let $\mathcal{A}$ be an abelian category with enough projective objects and of finite global dimension, and $B$ be a finite-dimensional algebra. If one has an equivalance of triangualted categories $D^\text{b}(\mathcal{A})\cong D^\text{b}(\text{mod}\ B)$, then $B$ is also of finite global dimension.
[1] Beilinson, A. A., Coherent sheaves on ({\mathbb{P}}^n) and problems of linear algebra, Funct. Anal. Appl. 12, 214-216 (1979). ZBL0424.14003.
We do know that all the algebras $B_n$ have finite global dimension since they are quiver algebras and the quiver has no oriented cycles. The projective resolution of each simple module over $B_n$ has length $n$ at most.
Actually, your statement about derived equivalences is also true. To compute the global dimension of $B$ we need to find the largest $n$ such that $\mathrm{Hom}_{D^b(B)}(X,Y[n])$ is nonzero for simple $X$ and $Y$. They are mapped to some bounded complexes over $\mathcal{A}$ under the given equivalence, which is of finite global dimension, so this $\mathrm{Hom}$ must also vanish for large enough $n$.