Constant Riemannian Metric Let $M=\mathbb R^n$ and define for each $x\in \mathbb R^n$, define 
$$\langle v,w\rangle_x= \langle v,w\rangle_0$$
where $v,w\in T_x\mathbb R^n\equiv \mathbb R^n\equiv T_0\mathbb R^n$. Hence we see that $M=\mathbb R^n$, admits constant metrics.
Is there any other manifold (other than $\mathbb R^n$) which admits constant metrics...
One guess may be Lie group admit constant metric. Is there any other example.   
 A: I'm interpreting your "admits constant metrics" in the following way: Any point $p\in M$ has a neighborhood $U$ with local coordinates $(u_i)_{1\leq i\leq n}$ such that in terms of the $u_i$ the metric tensor $g_{ik}$ is constant. This means that any point $p$ has a neighborhood $U$ which is isometric to a euclidean $n$-ball.
Examples of such manifolds can be produced in the following way: In $\Bbb R^n$, provided with the standard euclidean metric, take $r$ linearly independent vectors $a_1$, $\ldots$, $a_r$ and call two points $x$, $y\in\Bbb R^n$ equivalent if
$$y-x=\sum_{k=1}^r j_k \>a_k,\qquad j_k\in{\mathbb Z}\quad (1\leq k\leq r)\ .$$
It's easy to check that $M:=\Bbb R^n{/}\!\sim\ $ has the desired property. When $r=n$ then $M$ is compact and is called an $n$-dimensional flat torus.
A: I'm interpreting the question differently:  Namely, when can I use precisely what you did to define a metric?
The issue is that you can't always make a well defined identification of the tangent space at one point with the tangent space at another.  So, saying $v\in T_p M$ "corresponds" to $v'\in T_q M$ doesn't always make sense.
For example, on $S^2$, given a nonzero tangent vector at the north pole, it should be clear that if this "corresponds" to anything, it should "correspond" to something nonzero.  Then, a "correspondance" to every point on $S^2$ is nothing but a vector field on $S^2$ which vanishes nowhere.  Unfortunately, the Hairy Ball Theorem states that there is no such vector field.
The class of manifolds you are looking for are the so called Parallelizable Manifolds.  These are precisely those manifolds for which the tangent bundle admits a global trivialization.  After choosing such a trivialization, choosing any basis for the tangent space at some particular point.  Then, via the trivialaztion, this defines a basis at each point of your manifold and this gives you a way of identifying tangent spaces at different points.
I want to stress that this is very noncanonical - you must choose a trivialization and a basis.  Different choices will give rise to different Riemannian metrics with different properties.
Parallizable manifolds include $\mathbb{R}^n$ and all Lie groups, as you guessed, but also contain many more manifolds.  For example, $S^7$ is not a Lie group, but is parallelizable.  Further, the product of any sphere with an odd dimensional sphere is parallelizable.
