What can you see on a hexagonal spiral?
JavaScript (ES6),  178 169  168 bytes
Returns the \$n\$ first terms as an array.

n=&gt;(X=[1],Y=[L=1],d=5,j=x=y=0,g=a=&gt;--n?g([...a,(X[x+=~-'210012'[d%=6]]=-~X[x])+(Y[y+=~-'1221'[j?d:d++]]=-~Y[y])+(Z[x+y]=-~Z[x+y])-2],++j%L?0:(j%=L*6)?d++:L++):a)(Z=[1])

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How?
This is a rather naive approach that actually computes the spiral with cube coordinates \$(x,y,x+y)\$ and keeps track of the number of cells on each row \$Z[x+y]\$, each diagonal \$Y[y]\$ and each anti-diagonal \$X[x]\$.
Commented
n =&gt; (                            // n = input
  X = [1], Y = [L = 1],           // X[] = anti-diagonals, Y[] = diagonals, L = layer
  d = 5,                          // d = direction in [0..5]
  j =                             // j = cell index in the current layer
  x = y = 0,                      // (x, y) = coordinates
  g = a =&gt;                        // g = main recursive function taking the output a[]
    --n ?                         // decrement n; if we haven't computed enough terms:
      g(                          //   recursive call:
        [ ...a,                   //     append the previous terms
          ( X[x +=                //     update x and increment X[x]
              ~-'210012'[d %= 6]  //       pick dx according to d
            ] = -~X[x]            //
          ) +                     //
          ( Y[y +=                //     update y and increment Y[y]
              ~-'1221'[j ? d      //       pick dy according to d
                         : d++]   //       increment d afterwards if j = 0
            ] = -~Y[y]            //       (i.e. this is the 1st cell of the layer)
          ) +                     //
          (Z[x + y] = -~Z[x + y]) //     increment Z[x + y]
          - 2                     //     subtract 2 from X[x] + Y[y] + Z[z] because
        ],                        //     the current cell is counted three times
        ++j                       //     increment j
        % L ?                     //     if it's not the last cell of this side:
          0                       //       do nothing
        :                         //     else:
          (j %= L * 6) ?          //       if it's not the last cell of the layer:
            d++                   //         increment d
          :                       //       else:
            L++                   //         increment L
      )                           //   end of recursive call
    :                             // else:
      a                           //   we're done: return a[]
)(Z = [1])                        // initial call to g with a[] = Z[] = [1]