Even when computing $\pi_k(S^n), k < n$ all of the hard work, as far as I can tell, comes from showing that continuous maps behave reasonably up to homotopy; there is no difficulty once you show, whatever way you like (simplicial approximation, smooth approximation, etc.), that you can ignore space-filling curves. Until you know that it's necessary to consider the possibility that continuous maps behave in totally ungeometric ways that make them unsuitable for modeling homotopy theory.

Topological spaces and continuous maps are both absurdly general objects, and in fact too general to model homotopy theory: instead of working with the homotopy category of topological spaces we should of course be working with the homotopy category of CW complexes, and of course there are other ways to describe the homotopy category that don't involve topological spaces at all. In some of these descriptions (starting from simplicial sets, I would guess) it may be quite easy to compute $\pi_k(S^n), k \le n$.

Morally this computation should be easy starting from a heuristic picture of $\infty$-groupoids: $S^n$ is the free $\infty$-groupoid on an $n$-morphism, and in particular has no interesting $k$-morphisms for $k < n$. For $k = n$, again heuristically, all you can do starting from an $n$-morphism is compose it with itself a lot; the $\mathbb{Z}$ appearing here is the free group on one generator. Possibly this is the sort of reasoning that homotopy type theory is supposed to make precise.

Edit: That heuristic reasoning above may not seem too convincing because it doesn't seem to say anything about the higher homotopy groups, so let me spell out what it suggests about $\pi_3(S^2)$.

To compute this it suffices to ask what the free $3$-groupoid on a $2$-morphism is. A $3$-category with one object and one $1$-morphism is precisely a braided monoidal category, so the question now is what the free grouplike braided monoidal groupoid on an object $X$ is like. Well, it has a dual $X^{\ast}$ (which must in fact be its inverse) and there are unit and counit maps $1 \to X \otimes X^{\ast}, X^{\ast} \otimes X \to 1$. Finally there is a braiding $X \otimes X^{\ast} \to X^{\ast} \otimes X$. These three maps can be composed, and we get a map $1 \to 1$ which in no way is required to be the identity; in fact it can be identified with the generator of $\pi_3(S^2)$.

Incidentally, the corresponding argument for $\pi_2(S^1)$ fails because we don't have a braiding; the corresponding question is what the free grouplike monoidal groupoid on an object is like. The key is to be extremely careful what the unit and counit look like in a monoidal category which is not assumed to be symmetric, and once we are, the argument correctly suggests that $\pi_2(S^1)$ is trivial.

Even when computing $\pi_k(S^n), k < n$ all of the hard work, as far as I can tell, comes from showing that continuous maps behave reasonably up to homotopy; there is no difficulty once you show, whatever way you like (simplicial approximation, smooth approximation, etc.), that you can ignore space-filling curves. Until you know that it's necessary to consider the possibility that continuous maps behave in totally ungeometric ways that make them unsuitable for modeling homotopy theory.

Topological spaces and continuous maps are both absurdly general objects, and in fact too general to model homotopy theory: instead of working with the homotopy category of topological spaces we should of course be working with the homotopy category of CW complexes, and of course there are other ways to describe the homotopy category that don't involve topological spaces at all. In some of these descriptions (starting from simplicial sets, I would guess) it may be quite easy to compute $\pi_k(S^n), k \le n$.

Morally this computation should be easy starting from a heuristic picture of $\infty$-groupoids: $S^n$ is the free $\infty$-groupoid on an $n$-morphism, and in particular has no interesting $k$-morphisms for $k < n$. For $k = n$, again heuristically, all you can do starting from an $n$-morphism is compose it with itself a lot; the $\mathbb{Z}$ appearing here is the free group on one generator. Possibly this is the sort of reasoning that homotopy type theory is supposed to make precise.

Edit: That heuristic reasoning above may not seem too convincing because it doesn't seem to say anything about the higher homotopy groups, so let me spell out what it suggests about $\pi_3(S^2)$.

To compute this it suffices to ask what the free $3$-groupoid on a $2$-morphism is. A $3$-category with one object and one $1$-morphism is precisely a braided monoidal category, so the question now is what the free grouplike braided monoidal groupoid on an object $X$ is like. Well, it has a dual $X^{\ast}$ (which must in fact be its inverse) and there are unit and counit maps $1 \to X \otimes X^{\ast}, X^{\ast} \otimes X \to 1$. Finally there is a braiding $X \otimes X^{\ast} \to X^{\ast} \otimes X$. These three maps can be composed, and we get a map $1 \to 1$ which in no way is required to be the identity; in fact it can be identified with the generator of $\pi_3(S^2)$.

Incidentally, the corresponding argument for $\pi_2(S^1)$ fails because we don't have a braiding; the corresponding question is what the free grouplike monoidal groupoid on an object is like. The key is to be extremely careful what the unit and counit look like in a monoidal category which is not assumed to be symmetric, and once we are, the argument correctly suggests that $\pi_2(S^1)$ is trivial.

Edit #2: And the Freudenthal suspension theorem appears here in the fact that for $\pi_{n+1}(S^n), n \ge 3$ the question stabilizes to looking at the free grouplike symmetric monoidal groupoid on an object.

Even when computing $\pi_k(S^n), k < n$ all of the hard work, as far as I can tell, comes from showing that continuous maps behave reasonably up to homotopy; there is no difficulty once you show, whatever way you like (simplicial approximation, smooth approximation, etc.), that you can ignore space-filling curves. Until you know that it's necessary to consider the possibility that continuous maps behave in totally ungeometric ways that make them unsuitable for modeling homotopy theory.

Topological spaces and continuous maps are both absurdly general objects, and in fact too general to model homotopy theory: instead of working with the homotopy category of topological spaces we should of course be working with the homotopy category of CW complexes, and of course there are other ways to describe the homotopy category that don't involve topological spaces at all. In some of these descriptions (starting from simplicial sets, I would guess) it may be quite easy to compute $\pi_k(S^n), k \le n$.

Morally this computation should be easy starting from a heuristic picture of $\infty$-groupoids: $S^n$ is the free $\infty$-groupoid on an $n$-morphism, and in particular has no interesting $k$-morphisms for $k < n$. For $k = n$, again heuristically, all you can do starting from an $n$-morphism is compose it with itself a lot; the $\mathbb{Z}$ appearing here is the free group on one generator. Possibly this is the sort of reasoning that homotopy type theory is supposed to make precise.

Even when computing $\pi_k(S^n), k < n$ all of the hard work, as far as I can tell, comes from showing that continuous maps behave reasonably up to homotopy; there is no difficulty once you show, whatever way you like (simplicial approximation, smooth approximation, etc.), that you can ignore space-filling curves. Until you know that it's necessary to consider the possibility that continuous maps behave in totally ungeometric ways that make them unsuitable for modeling homotopy theory.

Topological spaces and continuous maps are both absurdly general objects, and in fact too general to model homotopy theory: instead of working with the homotopy category of topological spaces we should of course be working with the homotopy category of CW complexes, and of course there are other ways to describe the homotopy category that don't involve topological spaces at all. In some of these descriptions (starting from simplicial sets, I would guess) it may be quite easy to compute $\pi_k(S^n), k \le n$.

Morally this computation should be easy starting from a heuristic picture of $\infty$-groupoids: $S^n$ is the free $\infty$-groupoid on an $n$-morphism, and in particular has no interesting $k$-morphisms for $k < n$. For $k = n$, again heuristically, all you can do starting from an $n$-morphism is compose it with itself a lot; the $\mathbb{Z}$ appearing here is the free group on one generator. Possibly this is the sort of reasoning that homotopy type theory is supposed to make precise.

Edit: That heuristic reasoning above may not seem too convincing because it doesn't seem to say anything about the higher homotopy groups, so let me spell out what it suggests about $\pi_3(S^2)$.

To compute this it suffices to ask what the free $3$-groupoid on a $2$-morphism is. A $3$-category with one object and one $1$-morphism is precisely a braided monoidal category, so the question now is what the free grouplike braided monoidal groupoid on an object $X$ is like. Well, it has a dual $X^{\ast}$ (which must in fact be its inverse) and there are unit and counit maps $1 \to X \otimes X^{\ast}, X^{\ast} \otimes X \to 1$. Finally there is a braiding $X \otimes X^{\ast} \to X^{\ast} \otimes X$. These three maps can be composed, and we get a map $1 \to 1$ which in no way is required to be the identity; in fact it can be identified with the generator of $\pi_3(S^2)$.

Incidentally, the corresponding argument for $\pi_2(S^1)$ fails because we don't have a braiding; the corresponding question is what the free grouplike monoidal groupoid on an object is like. The key is to be extremely careful what the unit and counit look like in a monoidal category which is not assumed to be symmetric, and once we are, the argument correctly suggests that $\pi_2(S^1)$ is trivial.