Sum pyramid of primes

J, 15 bytes

p:@i.+/ .*i.!<:

Explanation:

Basically the same as my Mathematica answer.

p:@i.+/ .*i.!<:
          i.!<:    binomial coefficients
p:@i.              first n primes
     +/ .*         dot product

Mathematica, 38 36 35 bytes

Prime[r=Range@#].Binomial[#-1,r-1]&

Minkolang 0.14, 17 bytes

n[i3M$i1-i6M*+]N.

Try it here and check all test cases here.

Explanation

n                    Take number from input (N)
 [                   Open for loop that repeats N times
  i                  Loop counter (n)
   3M                Pop n and push nth prime (where 2 is the 0th prime)
     $i1-            Max iterations - 1 (which is N-1)
         i           Loop counter (n)
          6M         Pop n,k and push kCn (binomial)
            *+       Multiply and add
              ]      Close for loop
               N.    Output as number and stop.

I use basically the same algorithm as several of the earlier answers that use binomial coefficients. Whenever you see such a pyramid of numbers being added, Pascal's triangle should be the first thing to come to mind. I don't see that any of the other answers have explained why this works, so I'll do that.

MORE explanation

2
  > [2,3]
3         > [2,3,3,5]
  > [3,5]             > [2,3,3,3,5,5,5,7]
5         > [3,5,5,7]
  > [5,7]
7

As you can see, the primes 2,3,5,7 appear 1,3,3,1 times in the final result. Lemme change the layout a bit.

_ _ _ 7
_ _ 5
_ 3
2

The number of times that the 3 will contribute to the final result is the same as the number of paths from the 3 to the upper-left corner, moving only up and left. Here, there are three such paths for the 3:

_    _    _ _
_    _ _    _
_ 3    3    3

Note that I can reverse the direction without loss of generality. So I want to know how many paths there are from the upper-left corner to each position along the jagged edge. I can count them like so...

1 1 1 1 1 . . .
1 2 3 4
1 3 6
1 4   .
1       .
.         .
.
.

For every number in this triangle, if it is X units from the left and Y units from the top, then the number at that position is

The way I use it, though, X+Y = N is constant and X ranges from 0 to N, which goes along one diagonal. I multiply each coefficient with the corresponding prime number and then add it all up.

See the Wikipedia article on Pascal's triangle for more on this.