There is a closed non-zero $n$-form on $\text{GL}(n, \Bbb{R})$ How to Prove that

There is a closed non-zero $n$-form $\omega$ on $\text{GL}(n, \mathbb{R})$ which is left and right invariant.

 A: There is one when $n=1$ and there is not one when $n=2$.  In fact, when $n=2$, there are simply no nonzero left and right invariant $2$-forms, irrespectively of whether we're talking about closed forms or not.  I'm fairly certain this implies that there isn't one for any higher $n$, but the details are currently eluding me.
If $n=1$, $Gl(1)$ is $\mathbb{R}^\ast$, all nonzero real numbers (with multiplication as the group operation).  Since $Gl(1)$ is abelian, any left invariant 1-form is automatically right invariant.  Then, pick your favorite nonzero 1-form at any point and use the group multiplication to move it around to all the other points.  The resulting 1-form is left and right invariant.
What about $n=2$?  First, some general facts.  Let $G$ be a Lie group, and for any $g\in G$, let $L_g, R_g:G\rightarrow G$ denote left and right multiplication by $g$, that is $L_g(h) = gh$ and $R_g(h) = hg$.  Let $C_g(h) = gh^{-1}$ denote conjugation.
Lemma:  If $\omega$ is a left invariant form on $G$, then $\omega$ is right invariant iff $\omega(e)$ is conjugation invariant.
Proof (sketch):  If $\omega$ is right invariant, then $\omega(e)$ is conjugation since $C_g^\ast \omega(e) = (L_g\circ R_{g^{-1}})^\ast \omega(e) = R_{g^{-1}}^\ast L_g^\ast \omega(e) = \omega(e)$ by left and right invariance.
On the other hand, notice that for every $h\in G$ $R_{h}^\ast \omega$ is left invariant since $L_g$ and $R_h$ commute.  Hence so long as $(R_h^\ast \omega)(e) = \omega(e)$, they must be the same form.  But by left invariance we have $(R_h^\ast \omega)(e) = (L_{h^{-1}}^\ast R_h^\ast \omega)(e) = C_{h^{-1}}^\ast \omega (e) = \omega(e)$. $\square$
So, so show there is no left and right invariant 2-form on $Gl(2)$, we're reduced to showing there is no 2-form on $T_e GL(2) = \mathfrak{gl}(2)$ which is invariant under  conjugation by elements of $GL(2)$.
Identify $\mathfrak{gl}(2)$ with $M_2(\mathbb{R})$.  Let $E_{ij}$ denote the standard basis having a $1$ in the $ij^{th}$ position and $0$s elsewhere.  Let $E_{ij}^\ast$ denote the dual basis.
Then any 2-form has the form \begin{align*} \omega &= \alpha_{11,12}E_{11}^\ast\wedge E_{12}^\ast + \alpha_{11,21}E_{11}^\ast\wedge E_{21}^\ast + \alpha_{11,22}E_{11}^\ast \wedge E_{22}^\ast \\\ &+ \: \alpha_{12,21}E_{12}^\ast\wedge E_{21}^\ast + \alpha_{12,22}E_{12}^\ast \wedge E_{22}^\ast + \alpha_{21,22}E_{21}^\ast \wedge E_{22}^\ast\end{align*}
Now we'll trim it down a bit.  Let $A= \begin{bmatrix} \lambda & 0 \\ 0 & 1\end{bmatrix}$ where $\lambda \neq 0$.  Note that $AE_{11}A^{-1} = AE_{22}A^{-1}$ but $AE_{12}A^{-1} = \lambda E_{12}$ and $AE_{21}A^{-1} = \lambda^{-1} E_{21}$.  Then conjugation invariance implies that any term involving $E_{12}$ or $E_{21}$ cannot involve $E_{11}$ nor $E_{22}$.
Hence, conjguation invariance implies $\omega$ has the simpler form $$\omega = \alpha_{11,22}E_{11}^\ast\wedge E_{22} + \alpha_{12,21}E_{12}^\ast \wedge E_{21}^\ast.$$
Let's trim further.  Let $B = \begin{bmatrix}0&1\\1&0\end{bmatrix}$.  We have $BE_{11}B^{-1} = E_{22}$, $BE_{22}B^{-1} = E_{11}$, and likewise $B$ permutes $E_{12}$ and $E_{21}$.  Conjugation invariance now implies the last two terms are $0$.
So we've shown the only possibly left and right invariant 2-form on $GL(2)$ is the $0$ form (and we haven't even worried at all about closedness of the form!) $\square$
