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Marc Hoyois
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I'm not sure if that's exactly what you mean by "being motivic", but this long exact sequence comes from a triangle in Voevodsky's category of (integral) motives $DM(\mathbb Q)$.

More generally if $X$ is a $\mathbb Q$-scheme of finite type, $Z\subset X$ a closed subscheme, $\pi : Y \to X$ the blowup at $Z$, and $E=\pi^{-1}(Z)$, then there is a triangle $$ M(E) \to M(Y)\oplus M(Z) \to M(X) \to M(E)[1] $$ in $DM(\mathbb Q)$. This is proved in Mazza–Voevodsky–Weibel's Lecture notes on motivic cohomology, see equation (14.5.3). It can also be viewed a formal consequence of the six-functor formalism for $DM(-)$, as described by Cisinski and Déglise in their paper Integral mixed motives in equal characteristic.

Added later: Let me sketch the abstract proof, which works in the $\ell$-adic context as well. The proof only uses that $\pi:Y\to X$ is a proper map that restricts to an isomorphism over the complement of $Z$ (an "abstract blow-up"). The triangle comes from a homotopy cocartesian square $1_X=M(Y) \coprod_{M(E)}M(Z)$ in $DM(X)$. Here for $f: X' \to X$ of finite type, we define $M(X')=f_!f^!(1_X)$. Let $i: Z\to X$ be the inclusion and $j: U\to X$ the open complement. Then the pair of functors $(i^!,j^*)$ is conservative. But applying either functor to the given square, using the proper base change theorem, gives a square which is cocartesian for formal reasons (opposite sides are equivalences).

Marc Hoyois
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