If I understand your group correctly, your quotient $G_n = B_n / \langle T_kS_k : k = 1,2,\cdots, n\rangle$ can be thought of as a fairly geometric object -- it's the mapping class group of a sphere with $n$ marked points.
Let $\pi_0 Diff(S^2,n+1)$ be the mapping class group of $S^2$ that preserves $n+1$ points. There is an epi-morphism $\pi_0 Diff(S^2,n+1) \to \Sigma_{n+1}$, so we can talk about the subgroup of $\pi_0 Diff(S^2,n+1)$ that fixes one of the $n+1$ permuted points, call it $A_n$. I believe
$$ A_n \simeq B_n / \langle T_n \rangle$$
and I believe Ian Agol mentioned this. I'm using angle brackets to indicate "normal subgroup generated by".
But we still have to account for the other relators you want to mod out by, $T_1S_1, T_2S_2, \cdots, T_{n-1}S_{n-1}$.
Let $X_n$ be the subgroup of $Diff(S^2,n+1)$ that stabilizes one of the $n+1$ points. Then there is a map $X_n \to Diff(S^2,n)$ given by forgetting that point. This map is the fibre of a locally-trivial fibre bundle with base-space a $n$-times punctured sphere. Moreover, this expresses $\pi_0 Diff(S^2,n)$ as a quotient of $A_n$, and the relators are precisely the remaining ones you list (this requires a bit of a computation).
So I believe the above is an argument your group $G_n$ is $\pi_0 Diff(S^2,n)$. This explains why you get $\mathbb Z_2$ in the $n=2$ case, and $\Sigma_3$ in the $n=3$ case. This group is linear for all $n$. There's two published proofs of this, one by Bigelow and myself and another by Korkmaz. The representation is obtained as a tweaking of the Lawrence-Krammer representation.
Revision: For $n>6$, there is no embedding of $\mathcal{S}_n \hookrightarrow \mathcal{S}_{n+1}$.
First, recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$, where $Mod(S_{0,n})$ is the (orientation preserving) mapping class group of the $n$-punctured sphere.
Inside $Mod(S_{0,n})$, there is a subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$, which is a rotation of order $n-2$ of $S^2$, and fixes the north and south poles. The $n$ punctures include the north and south poles and one orbit of size $n-2$. The preimage of this group in $\mathcal{S}_n$ is isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ or to $\mathbb{Z}/2(n-2)\mathbb{Z}$ (of course if $n$ is odd, these are isomorphic). Then we get a corresponding subgroup of $\mathcal{S}_{n+1}$ from the injection. Project this group to $Mod(S_{0,n+1})$ via the map $\mathcal{S}_{n+1}\to Mod(S_{0,n+1})$. The kernel of this projection will have size at most $2$. By the Nielsen Realization Theorem, any finite subgroup of $Mod(S_{0,n+1})$ must preserve a complete hyperbolic metric of finite area on $S_{0,n+1}$, and in particular by uniformization extends to a finite group of conformal automorphisms of $S^2$ which permutes $n+1$ marked points. The finite subgroups of $PSL_2(\mathbb{C})$ lie inside a conjugate of $SO(3)$, so the image is an abelian subgroup of $SO(3)$. The only abelian subgroups of $SO(3)$ are cyclic (or $\mathbb{Z}/2\mathbb{Z}^2$), so the image is either $\mathbb{Z}/(n-2)\mathbb{Z}$ or $\mathbb{Z}/2(n-2)\mathbb{Z}$ (in which case we may take an index 2 subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$; here we need $n>6$ to conclude that the image is not $\mathbb{Z}/2^2$). However, for $n>5$, there is no subgroup of $SO(3)$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ which permutes $n+1$ points, and therefore there is no such subgroup of $Mod(S_{0,n+1})$, a contradiction. To see this, note that a cyclic group of rotations of $S^2$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ has two fixed points, and every other orbit of size $n-2$. Thus, there must be some $k$ and $e$ such that there are $k$ orbits of size $n-2$, and $e$ orbits of size $1$, where $e\leq 2$. If $k\leq 1$, then we get $n+1=k(n-2)+e \leq n$, a contradiction. If $k\geq 2$, then $n+1=k(n-2)+e \geq 2(n-2)$, so $n\leq 5$, a contradiction.
Best Answer
Following up what was mentioned in the comments for $n$ up to $5$. In "Factor groups of the braid group" Coxeter showed that the quotient of the Braid group by the normal closure of the subgroup generated by $\{\sigma_i^k \ | \ 1\le i\le n-1\}$ is finite if and only if $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}$$ In your case ($k=3$) this translates to this group being infinite for $n\geq 6$.
P.S. For the same question on Artin braid groups one can use the classification of finite complex reflection groups. See for example the first reference there, "On complex reflection groups and their associated braid groups" by Broué, Malle and Rouquier.