Solution 1:

Per your profile, this is so you can sleep at night. Your space is exactly $\ell^p$. Here is a proof:

Suppose that $\|a\|_p<\infty$ and $\|f_n\|_p\le |a_n|$ for each $n$. Then $\|f_n^p\|_1\le |a_n|^p$ for each $n$, so that $\|\sum |f_n|^p\|_1<\infty$. In particular, $\sum_n |f_n|^p$ is finite almost everywhere, so that $|f_n|^p$ converges to 0 almost everywhere and hence $f_n$ converges to 0 almost everywhere.

For the converse, suppose that $\|a\|_p=\infty$. Then we construct a sequence of functions $f_n$ on $[0,1]$ with $\|f_n\|_p\le a_n$ so that $f_n(x)$ does not converge for every $x\in [0,1]$. Let $b_n=\min(|a_n|^p,\frac 12)$ and notice that if $f_n$ is the indictator function of an interval of length $b_n$, then $\|f_n\|_p=b_n^{1/p}\le |a_n|$. By assumption, $\sum b_n=\infty$.

Now we just use the standard construction of a sequence of functions that converges to zero in $L^1$, but not pointwise. Let $t_0=0$ and $t_{n}=(t_{n-1}+b_n)\bmod 1$ for each $n$. Then let $f_n=\mathbf 1_{[t_{n-1},t_n]}$ (where if $t_n<t_{n-1}$, this means $\mathbf 1_{[t_{n-1},1]} + \mathbf 1_{[0,t_n]}$. Now it is known that $\limsup f_n(x)=1$ for all $x$ and $\liminf f_n(x)=0$ for all $x$. (This has been described to me as the "Goodyear blimp": the support of the function continually moves to the right, each one starting where the previous one finished. The blimp passes overhead infinitely often because the $b_n$ are not summable).