$\newcommand{tot}{\mathbb{C}^n\setminus 0}\newcommand{tan}{\mathcal{T}_{\mathbb{P}^{n-1}}}$ Since morphism $\pi$ is affine, for any quasicoherent sheaf $\mathcal{F}$ on $\mathbb{C}^n\setminus 0$ its higher direct images $R^{>0}\pi_*\mathcal{F}$ vanish, so from Leray spectral sequence we get $H^i(\tot, \mathcal{F})=H^i(\mathbb{P}^{n-1},\pi_*\mathcal{F})$. Applying this to $\pi^*\mathcal{T}_{\mathbb{P}^{n-1}}$ we get $$H^1(\tot,\pi^*\tan)=H^1(\mathbb{P}^{n-1},\pi_*\pi^*\tan)=H^1(\mathbb{P}^{n-1},\tan\otimes\pi_*\mathcal{O}_{\tot})$$ by projection formula.
Note that $\pi$ is the projection from the total space of $ \mathcal{O}_{\mathbb{P}^{n-1}}(-1)$ minus zero section.(the following is corrected thanks to Jason Starr). The direct image of the structure sheaf of the total space itself is $Sym(\mathcal{O}(-1))=\bigoplus\limits_{n\geq 0}S^n\mathcal{O}(-1)=\bigoplus\limits_{n\geq 0}\mathcal{O}(-n)$. So, if we throw away the zero section, sections of structure sheaf are allowed to have a pole along the zero section, so $\pi_*\mathcal{O}_{\tot}=\bigoplus\limits_{n\in\mathbb{Z}}\mathcal{O}(-n)$. So, the problem is reduced to computing $H^1(\mathbb{P}^{n-1}, \tan\otimes \mathcal{O}(-k))$. From Euler exact sequence we get a long exact seqeunce
$\dots\to H^1(\mathbb{P}^{n-1},\mathcal{O}(1-k)^{\oplus n})\to H^1(\mathbb{P}^{n-1},\tan\otimes \mathcal{O}(-k))\to H^2(\mathbb{P}^{n-1},\mathcal{O}(-k))\to \dots$
Both left and right groups are zero, because line bundles on $\mathbb{P}^{n-1}$ can have nonzero cohoomology only in degrees $0,n-1>2$ so $H^1(\tot,\pi^*\tan)=0$