Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$? 
Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.

The only reduction I thought about so far would be:
$$n\cdot 1 + (n - 1)\cdot 2 + (n - 3)\cdot 3 +\cdots $$
which seems symmetrical; for example, an odd and even n
 5 4 3 2 1        4 3 2 1
*                *
 1 2 3 4 5        1 2 3 4

but I'm not sure if and how it may help derive a formula.

I think the initial steps towards a formula look a bit simpler than seen so far here.
Beginning with
$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$
we count the number of occurences of the $1$, of the $2$
$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$
and rewrite a bit
$$ \begin{align} H^{(2)}(n) &= 1\cdot ((n+1)-1) + 2 \cdot ((n+1)-2) + 3\cdot ((n+1)-3) + ... + n \cdot ((n+1)-n) \\ \qquad \\
&= (1+2+3+...+n)(n+1) - (1^2+2^2+3^2+...+n^2) \\ \qquad \\
&= {n(n+1) \over 2}(n+1) - {2n^3 + 3n^2 + 1n\over 6} \end{align}$$
The last steps stems from the faulhaber formulae for sums-of-like-powers.

Final step, expanding & collecting this gives the same result as derived by other answerers:
$$ H^{(2)}(n) = {3n(n+1)(n+1)-2n^3 - 3n^2 - 1n \over 6} \\
= {n^3+3n^2+2n \over 6} \\
= {n(n+1)(n+2) \over 3!}  $$
For $n=5$ we get the sum $H^{(2)}(5)=35$
