We know that the flag variety $SL(2,\mathbb{C})/B$ which $B$ is Borel subgroup, can be identified with $\mathbb{P^1}$, What can we say about $SL(n,\mathbb{C})/B$ which $B$ is Borel subgroup of $SL(n,\mathbb{C})$

The variety $\mathrm{SL}(n,\mathbb{C})/B$ is the variety of complete flags in $\mathbb{C}^n$. That is, a point in the variety can be identified with a chain $$\{0\}=V_0\subset V_1\subset\dotsc\subset V_{n1}\subset V_n=\mathbb{C}^n$$ such that $\dim{V_i}=i$. This reduces to $\mathbb{P}^1$ in the case $n=2$, as the only nonproscribed datum at each point is the line $V_1\subset\mathbb{C}^2$. To prove this, generalizing the $n=2$ result, one checks that the "standard" Borel subgroup stabilizes the "standard" flag $$\{0\}\subset\langle e_1\rangle\subset\langle e_1,e_2\rangle\subset\cdots\subset\mathbb{C}^n$$ where the $e_i$ are the standard basis vectors. To expand a bit more, and give an answer that judging from the comments you might like better, let $P$ be a maximal parabolic subgroup of $\mathrm{SL}(n,\mathbb{C})$, so $P$ is proper, contains a Borel, and is maximal with respect to this property. Then $\mathrm{SL}(n,\mathbb{C})/P$ is the Grassmannian of $k$planes in $\mathbb{C}^n$, which more naturally generalizes $\mathbb{P}^1$ (the Grassmannian of $1$planes in $\mathbb{C}^2$). What's happening in the $n=2$ case is that $\mathrm{SL}(2,\mathbb{C})$ is so small that the only parabolic subgroup (up to conjugacy) is Borel. 

