Solution 1
The left translation maps have the form $L_a:\mathbb R^n \rightarrow \mathbb R^n: x\mapsto a+x$. So $D L_a(x)=\operatorname{id}$ for all $a,x\in\mathbb R^n$ ($D L_a$ is the total derivative of $L_a$). So the set of all left-invariant vector fields is the set of all constant vector fields $\mathfrak g=\{f:\mathbb R^n\rightarrow T\mathbb R^n: f = \operatorname{const}\}$ (which is isomorph to $\mathbb R^n$ because $\mathfrak g$ is $n$-dimensional).
Because $X\in \mathfrak g$ is constant, one has $[X,Y]=\mathcal L_X(Y)=0$ for any given vector field $Y$.
Solution 2
$(\mathbb R^n, +)$ can be embedded into $GL_n$ via $$f:\mathbb R^n \rightarrow GL_n: (x_1,x_2,\ldots,x_n) \mapsto \left(\begin{matrix} e^{x_1} & & & & \\ & e^{x_2} & & \\ & & \ddots & \\ & & & e^{x_n}\end{matrix}\right)$$ $f$ is well defined because the determinant $e^{x_1 + x_2 + \ldots + x_n}$ is always positive and because of $e^{a+b}=e^a\cdot e^b$ the function $f$ is a group homomorphism.
Via $f$ one can show that $(\mathbb R^n,+)$ is isomorph to the set $D^+$ of diagonal matrices with positive entries on the diagonal. The lie algebra $\mathfrak g$ of $D^+$ is given by $$\mathfrak g = \{\dot\gamma(0): \gamma:(-\epsilon,\epsilon)\rightarrow D^{+}\}$$ which is also $D^{+}$ (as one can easily show). Because $D^{+}$ is in the center of $GL_n$ the lie bracket is always zero.
The "scary expression" will give you what you want, but you need to be careful with the names of the indices, you have some conflicts there, like $i$ appearing twice in not summed expressions.
Instead, let's streamline things a bit, for $A,B\in \mathfrak g$ we have $$ A=a^{ij}\partial_{ij}|_e,\ B=b^{ij}\partial_{ij}|_e $$ with the components being constants, and summing over repeated indices are understood. A generic group element is denoted as $x^{ij}$ and $\partial_{ij}$ are the holonomic frame vectors associated with the canonical coordinate system mapping group elements to their matrix elements.
We also recall that given any manifold $M$ with local chart $(U,\varphi)$ (with $\varphi(x)=(x^1(x),...,x^n(x))$), if two vector fields $X=X^i\partial_i$ and $Y=Y^i\partial_i$ are given, then their commutator is locally given by $$ [X,Y]=\left(X^j\partial_j Y^i-Y^j\partial_j X^i\right)\partial_i. $$
Left multiplication:
Let $\gamma:(-\epsilon,\epsilon)\rightarrow G$ be a smooth curve such that $\gamma(0)=x$ and $\dot\gamma(0)=X=X^{ij}\partial_{ij}|_x$. Let $g=(g^{ij})\in G$ be a group element. Then $$ (l_g)_\ast X=\frac{d}{dt}g\gamma(t)|_{t=0}=^!gX=g^{ik}X^{kj}\partial_{ij}|_{gx}, $$ where at the equality sign with the exclamation mark we use the fact that the group elements are just ordinary matrices embedded into $\mathbb R^{n\times n}$.
So left translation of vectors in $\text{GL}(n,\mathbb R)$ is just ordinary left multiplication of the vector (which is a matrix, rememeber!).
The derivation:
By the previous, the left invariant vector fields corresponding to $A,B\in\mathfrak g$ (also denoted the same way) are given by $$ A_x=x^{ik}a^{kj}\partial_{ij}|_x\ B_x=x^{ik}b^{kj}\partial_{ij}|_x. $$
If we now reinterpret the $x^{ij}$ from being specific variables to being coordinate functions, we can also write the vector fields without evaluation at a specific point as $$ A=x^{ik}a^{kj}\partial_{ij}\ B=x^{ik}b^{kj}\partial_{ij}. $$ the commutator is then $$ [A,B]=(A^{mn}\partial_{mn}B^{ij}-B^{mn}\partial_{mn}A^{ij})\partial_{ij}, $$ where $$ A^{ij}=x^{ik}a^{kj}, $$ and similarly for $B$. This is $$ [A,B]=\left( x^{mr}a^{rn}\partial_{mn}(x^{ik}b^{kj})-x^{mr}b^{rn}\partial_{mn}(x^{ik}a^{kj}) \right)\partial_{ij}=^!\left(x^{mr}a^{rn}\delta^i_m\delta^k_n b^{kj}-x^{mr}b^{rn}\delta^i_m\delta^k_n a^{kj}\right)\partial_{ij} \\ =\left( x^{ir}a^{rk}b^{kj}-x^{ir}b^{rk}a^{kj} \right)\partial_{ij}. $$ At the equality with the exclamation mark we have used that $\partial_{mn}x^{ij}=\delta^i_m\delta^j_n$ and that the coefficients $a^{ij},b^{ij}$ are constants.
For the identity element we have $x^{ij}(e)=\delta^{ij}$, so $$ [A,B]_e=\left(a^{ik}b^{kj}-b^{ik}a^{kj}\right)\partial_{ij}|_e=[A_e,B_e], $$ where the last expression is the ordinary matrix commutator of the matrices $A_e,B_e$.
Best Answer
If $G$ and $H$ are Lie groups, if their Lie algebras are $\mathfrak g$ and $\mathfrak h$ respectively, and if $f\colon G\longrightarrow H$ is a Lie group homomorphism, them $f_e'\colon\mathfrak{g}\longrightarrow\mathfrak h$ is a Lie algebra homomorphism. Now, apply this to $G$ and to $GL(n,\mathbb{C})$. In your case, $f$ is the inclusion, and therefore $f_e'$ is the inclusion map too.