Here is an argument showing that adding a single Cohen reals yields an uncountable strong measure zero set.

In the spirit of the previous comments, let ${q_n}$ enumerate the rationals and for any function $f:\omega\to\mathbb{N}$ let $S(f) = \bigcup_{n\in\omega}[q_{f(n)},q_{f(n)}+1/f(n)]$. Let $B$ be the set of all reals $x$ such that for all $j\in \mathbb N$ there is $k$ such that $0< x- q_{j+k} < 1/(j+k)$. Then $B = \bigcap_{n\in\omega}S(n)$ and hence $B$ is a dense $G_\delta$ and it suffices to show that $V\cap B$ has strong measure zero after adding a Cohen real.

For this to be true it suffices to show that in the model obtained by adding a single Cohen real, for every function $f:\omega\to\mathbb{N}$ there is $g:\omega\to\mathbb{N}$ such that $g(n)\geq f(n)$ for all $n$ and $S(g)$ contains $V\cap B$.

To see that this is the case think of Cohen forcing as consisting of functions $p:n\to \mathbb{N}$ where $n\in\omega$ and think of the generic set $G$ as a function $G:\omega\to\mathbb{N}$. Begin by noting that if $f$ belongs to the ground model then $S(f+ G)$ contains $V\cap B$ because, given any Cohen condition $p$ with domain $n$ and any $x\in V\cap B$ there is $k$ such that $x\in [q_{f(n)+k}, q_{f(n)+k} + 1/(f(n) +k)]$ and extending $p$ to have value $k$ at $n$ yields a condition $q$ forcing $x\in S(f+ G)$.

On the other hand, if $f:\omega \to \mathbb{N}$ does not belong to $V$ then define $\bar{f}:\omega \to \mathbb{N}$ as follows: $\bar{f}(n)$ is the least $k$ such that there is $p\in G$ and $s\in \mathbb N$ such that $p:k\to \mathbb N$ and $p\Vdash f(\check{n}) =\check{s}< \check{k}$. It is clear that $f(n) < \bar{f}(n)$ for all $n$ so it suffices to show that $S(\bar{f}+ G)$ contains all ground model reals.

To this end, let $x\in B$ belong to the ground model and let $p:n\to \mathbb N$ be an arbitrary Cohen condition and suppose that infinitely many values of $f$ are not decided by $p$. Let $m\geq n$ be such that $p$ does not decide the value of $f(\check{m})$ and let $q\supseteq p$ be of minimal length such that $q\Vdash f(\check{m}) = \check{s}$ for some $s\in \mathbb N$. If $k$ is the domain of $q$ then $q\Vdash \bar{f}(\check{m}) = \check{k}$. (Why? No restriction of $p$ decides the value of $f(\check{m})$ so the minimality of $k$ implies that no restriction of $q$ decides the value of $f(\check{m})$ either.) Now use that $x\in V\cap B$ and argue as in the case when $f\in V$ to find $t$ such that that $x\in [q_{s+t},q_{s+t} + 1/(s+t)]$. Then extending $q$ to have value $t$ at $k$ yields a condition $\bar{q}$ forcing $x\in S(\bar{f}+ G)$.

Here is an argument showing that adding a single Cohen reals yields an uncountable strong measure zero set.

In the spirit of the previous comments, let $r_n$ for $n\in\omega$ enumerate the rationals. Let Cohen forcing be represented by the set of finite partial functions from $\omega$ to $\omega$ under inclusion and let $G$ be a name for the generic function $G:\omega \to \omega$. For any pair of functions $f:\omega \to \mathbb N$ and $F:\omega \to \omega$ define $B(f,F)=\bigcup_{n\in\omega}(r_{G(F(n))} - 1/f(n),r_{G(F(n))} + 1/f(n))$. It suffice to show that for any Cohen name $\dot{f}$ there is $F:\omega \to \omega$ such that it is forced that the ground model reals are contained in $B(\dot{f},F)$.

To this end, for each Cohen condition $p$ define $F_p(m)$ to be some integer $k$ such that there is $q:k\to \omega$ such that $q\supset p$ and $q$ decides a value for $\dot{f}(m)$. Let $F:\omega \to \omega$ be such that $F\geq^* F_p$ for each $p$. To see that it is forced that $B(\dot{f},F)$ contains the ground model reals let $p$ be an arbitrary Cohen condition and $x$ a ground model real. Let $m$ be such that $F(m)>F_p(m)$. Choose $q:k \to \omega$ witnessing that $F_p(m) = k$ and suppose that $q$ forces $\dot{f}(m)=j$. Let $i\in\omega$ be such that $x\in (r_i- 1/j,r_i+1/j)$. Let $q'= q\cup\{(F(m),i)\}$. Then $q'$ forces that $G(F(m))=i$ and hence that $x\in (r_{G(F(m))}- 1/\dot{f}(m),r_{G(F(m))}+1/\dot{f}(m))\subseteq B(\dot{f},F)$.