Coefficients of $(1+x+\dots+x^n)^3$? Consider the following polynomial:
$$ (1+x+\dots+x^n)^3 $$
The coefficients of the expansion for few values of $n$ ($n=1$ to $5$) are:
$$ 1, 3, 3, 1 $$
$$ 1, 3, 6, 7, 6, 3, 1 $$
$$ 1, 3, 6, 10, 12, 12, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1 $$
$$ 1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1 $$
Is there a closed-form formula for the $i$th element of this sequence (for different values of $n$)? 
Edit This looks similar to the sequence A109439 on OEIS corresponding to the coefficients of the expansion of: $ \left( \frac{1 - x^n}{1 - x} \right)^3 .$
 A: An alternative approach:
Just as the $x^i$ term in $(1+\dots+x^n+\dots)^3$ counts the number of ways of picking three numbers to add to $i$,  the $x^i$ term in your expansion counts the number of non-negative integer solutions to the system
$$y_1+y_2+y_3 = i$$
$$y_1, y_2, y_3 \leq n$$
If it wasn't for that pesky second condition we'd have a classic problem that can be solved by the Stars and Bars method (or whatever other name you wish to call it).  There's a total of $\binom{i+2}{2}$ solutions.  In fact, that second condition disappears once $n \geq i$, which explains why each term in your sequence eventually stabilizes (and matches the OEIS sequence you found).
We now need to subtract off solutions where some of the $y_i$ are too large.  To avoid overcounting, we use Inclusion-Exclusion, which tells us that the number we want is equal to
$$\binom{i+2}{2}-3S_1+3S_2-S_3,$$
where $S_1$ is the number of solutions with $y_1>n$ (the $3$ comes from the fact that we also count $y_2>n$ and $y_3>n$), $S_2$ is the number of solutions with $y_1>n$ and $y_2>n$, and $S_3$ counts the number of solutions with all three variables too large.
Now we use one final trick.  The number of solutions to $y_1+y_2+y_3=n$ with $y_1>n$ is the same as the number of solutions to
$$(y_1-(n+1))+y_2+y_3=i-(n+1)$$
$$y_1-(n+1), y_2, y_3 \geq 0$$
which is just $\binom{i-(n+1)+2}{2}=\binom{i+1-n}{2}$.  Doing an identical trick with $S_2$ and $S_3$ gives a final answer of
$$\binom{i+2}{2}-3\binom{i+1-n}{2}+3\binom{i-2n}{2}-\binom{i-1-3n}{2}.$$
(correspondence fixed thanks to Thomas's helpful correction from his comment)
A: Denote by
$$\sum_{k=0}^{3n}{\binom{3}{k}}_{n+1}x^{k}=(1 + x + ... + x^{n})^3\,$$ 
the expansion of multinomial of degree $n$
From that relation follow that 
$$ \sum_{k=0}^{3n)}{\binom{3}{k}}_{n+1}x^{k}=\left(\frac{1-x^{n+1}}{1-x}\right)^{3}=(1-x^{n+1})^{3}(1-x)^{-3}\,$$ 
from binomial formula we have
$$(1-x^{n+1})^{3}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}x^{(n+1)i}$$ 
and from Taylor formula 
$$(1-x)^{-3}=\sum_{j=0}^{\infty}\binom{2+j}{j}x^{j}$$ 
after substitutions we get 
$$ \sum_{k=0}^{\infty}{\binom{3}{k}}_{n+1}x^{k}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}x^{(n+1)i}\sum_{j=0}^{\infty}\binom{2+j}{j}x^{j}=$$ 
$$=\sum_{j=0}^{\infty}\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+j}{j}x^{(n+1)i+j}=$$ 
$$ =\sum_{k=0}^{\infty}\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+k-(n+1)i}{2}x^{k}\,$$ 
then equate the coefficients next to x follow formula 
$${\binom{3}{k}}_{(n+1)}=\sum_{i=0}^{3}(-1)^{i}\binom{3}{i}\binom{2+k-(n+1)i}{2}\,$$ 
formula for computing the coefficients of expansion
A: If the polynomial were replaced by an infinite sum, the coefficient of $x^i$ would be equal to the number of ways to choose $(a,b,c) \in \{0,1,2,...\}^3$ such that $a+b+c=i$.  This is equal to the number of ways to choose two numbers $\le i$, or $\frac{1}{2}(i+1)(i+2)$.  The only difference here is that you want to limit the count to cases where $a,b,c \le n$.  Using the inclusion-exclusion principle, you can subtract the cases where one of the numbers is greater than $n$, add back the cases where two of the numbers are greater than $n$, and subtract the cases where all three are greater than $n$.  For each of these counts, subtract $n+1$ (or $2n+2$ or $3n+3$) from the target sum first, returning you to the case where all three numbers are unrestricted.  The result is
$$
\begin{eqnarray}
a_{i,n}&=&\frac{1}{2}(i+1)(i+2) - \frac{3}{2}(i-n)(i-n+1)\Theta(i-n-1) \\
&=& + \frac{3}{2}(i-2n-1)(i-2n)\Theta(i-2n-2) - \frac{1}{2}(i-3n-2)(i-3n-1)\Theta(i-3n-3),
\end{eqnarray}
$$
where $\Theta(x)=1$ for $x\ge 0$ and $0$ for $x<0$.
