By "higher dimensonal class field theory", Tate means the class field theory of higher dimensional local fields (see also this brief discussion), developed in the work of various people, including Kato and Parshin.
As for your second question, about learning CFT from a modern perspective: with my own students, I encourage them to learn from Cassels and Frolich (including the exercises), from Cox's book Primes of the form $x^2 + n y^2$, and from Washington's article on Galois cohomology in Cornell, Silverman, and Stevens.
The first reference (especially the main articles of Serre and Tate) gives a development of the main results of CFT which I think is hard to beat. Cox's book gives an important classical perspective. Washington's article gives insight into how class field theory can be reformulated as a collection of theorems (mainly due to Tate) on (local and global) Galois cohomology.
Tate's article Number theoretic background in the second volume of Corvalis is good when you have reached a certain level of sophistication, and are ready to move on from just focussing on algebraic number theory and CFT to a broader perspective. My experience is that it is a little austere for a beginner, though.
One thing that you will be missing if you follow the above references is an $L$-function-based perspective on class field theory. I gather that this is discussed in the new edition of Artin--Tate. If so, it is worth learning, since although it is the more old-fashioned point of view on CFT, non-abelian class field theory (i.e. the Langlands program) is founded on the notion of $L$-functions. (I believe that Lang's book also discusses the $L$-function approach to CFT, but I've never read it myself.)
There are a few ways to generalize quadratic reciprocity. Let me describe for you how class field theory over $\mathbb{Q}$ works.
Let $K$ be a finite abelian extension of $\mathbb{Q}$ of degree $n$, Galois group $\Gamma$. One goal of class field theory is the following: give a general rule describing how prime numbers $p$ decompose into a product of primes in $\mathcal O_K$. It's cumbersome to give a succinct rule which describes how all primes split, so the usual workaround is to give a rule which applies to all primes except those in a finite $S$, usually containing all the ramified primes.
This generalizes quadratic reciprocity in the following way: let $K = \mathbb{Q}(\beta)$ for some $\beta \in \mathcal O_K$, and let $f \in \mathbb{Z}[X]$ be the minimal polynomial of $f$ over $\mathbb{Q}$. For almost all primes $p$, you have $(\mathcal O_K)_{(p)} = \mathbb{Z}_{(p)}[\beta]$, and so you can apply the following criterion: if you factor $f$ into a product of irreducibles $p_1^{e} \cdots p_g^{e}$ over $\mathbb{Z}/p\mathbb{Z}[X]$, then $e$ will be the ramification index of $p$ in $K$, the degree $f$ of any of the $p_i$ will be the inertia, and $p$ will split into $\frac{n}{ef}$ primes.
So splitting of primes is analogous to factoring certain polynomials over $\mathbb{Z}/p\mathbb{Z}[X]$. In particular, quadratic reciprocity deals with determining for which primes $p$ the polynomial $X^2 - q \in \mathbb{Z}/p\mathbb{Z}$ is irreducible for a fixed prime $q$ (possibly $q$ is a negative prime). In other words, quadratic reciprocity answers the question of which primes $p$ split in $\mathbb{Q}(\sqrt{q})$, or more generally which primes split in a quadratic extension of $\mathbb{Q}$.
Now, here is how class field theory gives you an algorithm for determining (with a finite number of exceptions) how primes split in $K$, or at least implies the existence of such an algorithm. Given a prime number $p$, fix a prime $\mathfrak p$ of $K$ lying over $p$. Remember that for $p$ unramified in $K$, the decomposition group $$\Gamma_{p} = \{ \sigma \in \Gamma : \sigma \mathfrak p = \mathfrak p \}$$ is cyclic, and it has a particularly nice generator, commonly denoted $(p, K/\mathbb{Q})$. It does not depend on the choice of $\mathfrak p$, because $K/\mathbb{Q}$ is abelian. It is called the Frobenius at $p$. The order $f = f_p$ of $(p,K/\mathbb{Q})$ is the inertial degree of $p$ in $K$, and $p$ splits into $\frac{n}{f}$ primes in $K$.
So given a divisor $g$ of $n$, we want an algorithm for determining which primes $p$ split into $g$ primes in $K$, i.e. which primes satisfy $g = \frac{n}{f_p}$.
The nonconstructive part of the proof is the Kronecker-Weber theorem. It says that there exists some integer $m$ such that $K$ is contained in $\mathbb{Q}(\zeta_m)$. Suppose we have found such an $m$. Let $L = \mathbb{Q}(\zeta_m)$, $G = \textrm{Gal}(L/\mathbb{Q})$, and $H = \textrm{Gal}(L/K)$. There is a canonical identification of $G$ with the group $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and under this identification we can think of $H$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$. For $p$ not dividing $m$, the Frobenius $(p,L/\mathbb{Q})$ has the effect $\zeta_m \mapsto \zeta_m^p$. In particular, $(p,L/\mathbb{Q})$ can be identified with $p$ modulo $m$, and the inertial degree of $p$ in $L$ is then the multiplicative order of $p$ modulo $m$.
In general, if $\phi: A \rightarrow B$ is a homomorphism of finite groups, and $a \in A$, then the order of $\phi(a)$ is the smallest number $f$ such that $a^f \in \textrm{Ker } \phi$. Now, you can check that the restriction of $(p,L/\mathbb{Q})$ to an automorphism of $K$ is exactly $(p,K/\mathbb{Q})$. Therefore, the inertial degree of $(p,K/\mathbb{Q})$ is the smallest number $f$ such that $p^f$ is congruent modulo $m$ to a member of $H$.
So to summarize, here is how to determine how primes split in $K$:
Find an $m$ such that $K \subseteq \mathbb{Q}(\zeta_m)$.
Identify the Galois group of $\mathbb{Q}(\zeta_m)$ with $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and identify $H = \textrm{Gal}(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$.
With the exception of the prime factors of $m$, the primes $p$ which split into $g$ factors in $K$ are exactly those primes for which $p+H$ has order $\frac{n}{g}$ in $G/H$.
Best Answer
To derive quadratic reciprocity from Artin reciprocity, consider the field extensions $$\mathbb{Q} \subset F=\mathbb{Q}(\sqrt{p^*}) \subset K=\mathbb{Q}(\zeta)$$ where $p$ is a prime $\ne 2$, $\zeta$ is a primitive $p$th root of unity, and $p^* = (-1)^{(p-1)/2} p$.
(To see that $F$ is contained in $K$, look up Gauss sums.)
Let $q$ be another prime, $q \ne p, q \ne 2$. We know that $p^*$ is a square mod $q$ iff $q$ splits in $F$ iff the Artin symbol of $q$ in $F/\mathbb{Q}$ is trivial. (All Galois groups considered here are abelian, so the Artin map depends only on the base prime $q$ and not on any particular prime in the upper field.)
The Artin symbol of $K/\mathbb{Q}$ over the prime $q$ is the element $\sigma_q: \zeta \mapsto \zeta^q$ in the Galois group of $K/\mathbb{Q}$.
But Artin($F/\mathbb{Q}$) is the restriction of Artin($K/\mathbb{Q}$) to $F$, so Artin($F/\mathbb{Q}$) is trivial iff $\sigma_q$ is in the kernel of this restriction map. Both Galois groups are cyclic, and so this restriction map is the unique nontrivial homomorphism from $(\mathbb{Z}/p\mathbb{Z})^{\times}$ to $\{\pm 1\}$. It is easy to check that $\sigma_q$ will be in the kernel of the restriction map iff $q$ is a square mod $p$.
Conclusion: $p^*$ is a square mod $q$ iff $q$ is a square mod $p$. This is quadratic reciprocity.
For other reciprocity laws, you'll have to choose the field $F$ differently. For example, for cubic reciprocity, choose $F$ to be the unique degree 3 extension in $K/\mathbb{Q}$. (Of course, this only exists if $p \equiv 1$ (mod 3), but if that condition fails then cubic reciprocity is not very interesting: everything will be a cube mod $p$.) Things get more complicated because the analog of the step "$p^*$ is a square mod $q$ iff $q$ splits in $F$" is not as simple. But it's the same idea.
Edit: This last paragraph is not quite correct. The proper extensions to consider for cubic reciprocity is $L \subset L(p^{1/3})$ where $L = \mathbb{Q}(\omega)$, $\omega$ a primitive cube root of unity. The Artin conductor of this extension is a divisor of $3p$, and given a prime $q \equiv 1$ (mod 3) in $L$ (relatively prime to $3p$), the image of the ideal $(q)$ given by Artin reciprocity is essentially the unique nontrivial map from the cyclic group $(O_L / q)^{\times}$ to the cyclic group Gal($L(p^{1/3})/L$) of order 3. For details, see the senior thesis of Noah Snyder referenced by KCd in the comments above. For general $n$th power reciprocity, use $L \subset L(p^{1/n})$ where $L$ is the cyclotomic field generated by a primitive $n$th root of unity.