N-dimensional Circles!

Mathematica, 18 bytes, up to ~168.15 trillion dimensions

Pi^(a=.5#)/a!#2^#&

Anonymous function. Takes two numbers as input, and returns an inexact number as output. Works with any number of dimensions. Outputs 1. for n = 0. Uses the formula from Volume of an n-ball on Wikipedia.

Explanation

We are attempting to compute π^n/2/Γ(n/2 + 1)·Rⁿ, or N[Pi^(n/2)/Gamma[n/2 + 1] R^n] in Mathematica. In our case, # (first argument) is n and #2 (second argument) is R. This leaves us with N[Pi^(#/2)/Gamma[#/2 + 1] #2^#] &, which can be golfed as follows:

N[Pi^(#/2)/Gamma[#/2 + 1] #2^#] &
Pi^(.5#)/Gamma[.5# + 1] #2^# &    (* replace exact with approximate numbers*)
Pi^(.5#)/(.5#)! #2^# &            (* n! == Gamma[n + 1] *)
Pi^(a=.5#)/a! #2^# &              (* replace repeated .5# *)
Pi^(a=.5#)/a!#2^#&                (* remove whitespace *)

and thus, our original program.

Jelly, 13 bytes + extra swag

÷2µØP*÷!
ç×*@

Try it online!

Works for any dimension, so long as the fixed value of π yielded by ØP (3.141592653589793) is accurate enough.

How?

÷2µØP*÷! - Link 1: n, r
÷2       - n / 2
  µ      - monadic chain separation
   ØP    - π (3.141592653589793)
     *   - exponentiate: π^(n/2)
       ! - Pi(n/2): Gamma(n/2 + 1)
      ÷  - divide: π^(n/2) / Gamma(n/2 + 1)

ç×*@     - Main link: n, r
ç        - call last link (1) as a dyad: π^(n/2) / Gamma(n/2 + 1)
  *@     - exponentiate with reversed @rguments: r^n
 ×       - multiply: r^n * π^(n/2) / Gamma(n/2 + 1)

R, 75 40 38 bytes (plus extra swag)

Well, looks like I could golf this down by giving in and using the gamma function rather than recursive functions.

function(n,r)pi^(n/2)/gamma(n/2+1)*r^n

Defines an anonymous function to calculate the volume of an n-dimensional hypersphere of radius r.

Some examples:

1 1 -> 2

0 48 -> 1

2 3 -> 28.27433

3 4.5 -> 381.7035

7 7 -> 3891048

100 3 -> 122051813

Swagless solution, 38 34 bytes

For a few bytes less, you can have an anonymous function that only works for dimensions 1 to 3. Returns numeric(0) for n=0, and NA for n>3. (numeric(0) is a numeric vector of length 0; NA is for "not available".) Performance is otherwise identical to the general solution above.

function(n,r)c(1,pi,4/3*pi)[n]*r^n