What does $GL_n(R)$ look like? Exactly as in the title - what does the general linear group "look like" (you are free to interpret this however you like) as submanifold of $R^{n^2}$? What should I imagine when I think of it? 
(I am aware that the simplest nontrivial case is already beyond visualization...)
 A: It's a big open subset. The set of singular matrices forms a conical hypersurface (with cone point at the origin).
You can see this quite explicitly for $GL(2)$, which is the complement of the hypersurface $x_1x_4-x_2x_3=0$ in $\Bbb R^4$.
A: I will ignore the submanifold part of the question and just concentrate on the question of what the general linear group looks like intrinsically.
Via Gram-Schmidt, $\text{GL}_n(\mathbb{R})$ is homotopy equivalent to the orthogonal group $\text{O}(n)$. In fact an even stronger statement is true, although I don't recall the proof: $\text{GL}_n(\mathbb{R})$ is diffeomorphic to $\text{O}(n)$ times a Euclidean space, so really to understand its topology it suffices to understand the topology of $\text{O}(n)$. Furthermore, $\text{O}(n)$ always consists of two diffeomorphic connected components, namely the two cosets of $\text{SO}(n)$, so to understand its topology it suffices to understand the topology of $\text{SO}(n)$.
For $n = 2$ things are particularly straightforward: $\text{SO}(2)$ is just the circle $S^1$. 
For $n = 3$ things are still not so bad: $\text{SO}(3)$ is just $\mathbb{RP}^3$, which is double covered by $\text{Spin}(3) \cong \text{SU}(2) \cong S^3$. One way to visualize $\mathbb{RP}^3$ is as the unit tangent bundle of $S^2$, so a point in $\mathbb{RP}^3$ can be thought of as a unit vector on the $2$-sphere: I discuss this in this blog post.
For $n = 4$ things are still not terribly bad: $\text{SO}(4)$ is double covered by $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2) \cong S^3 \times S^3$. $S^3$ itself can be visualized as $\mathbb{R}^3$ together with a point at infinity.
For $n \ge 5$ I only know how to make analogous statements up to rational homotopy equivalence. It's a general fact that any connected compact Lie group $G$ is rationally homotopy equivalent to a product of odd spheres; above those spheres were respectively $S^1, S^3$, and $S^3 \times S^3$. The general pattern depends on the parity of $n$ and I believe it goes like this: for $n = 2k$ even, $\text{SO}(n)$ is rationally homotopy equivalent to 
$$\left( S^3 \times S^7 \times \dots \times S^{4k-5} \right) \times S^{2k-1}$$
and for $n = 2k+1$ odd, $\text{SO}(n)$ is rationally homotopy equivalent to
$$S^3 \times S^7 \times \dots \times S^{4k-1}.$$
In particular, $\text{SO}(5)$ is rationally homotopy equivalent to $S^3 \times S^7$ and $\text{SO}(6)$ is rationally homotopy equivalent to $\left( S^3 \times S^7 \right) \times S^5$. 
For $n \ge 5$ it's also useful to know about the fiber sequences
$$\text{SO}(n-1) \to \text{SO}(n) \to S^{n-1}$$
coming from the action of $\text{SO}(n)$ on $S^{n-1}$. In general you should think of a fiber sequence $F \to E \to B$ as exhibiting the total space $E$ as a "twisted product" of the fiber $F$ and the base space $B$, so inductively the fiber sequences above allow you to exhibit all of the $\text{SO}(n)$s as "iterated twisted products of spheres." Unfortunately you get different spheres this way than the above way since some of them will be even spheres. 
The fiber sequences above also tell you that for fixed $i$, the homotopy groups $\pi_i(\text{SO}(n))$ stabilize as $n \to \infty$. The stable values are the homotopy groups of the stable (special) orthogonal group $\text{SO}$ and they are the subject of (real) Bott periodicity. 
