(An attempt at an answer, and also my first posting here.  My browser has problems with rendering, so I write in pseudo-TeX.)

I claim that a single Cohen real makes the set of old reals strong measure zero. 
Reals are functions from omega to 2. 

Let C be Cohen forcing, and let c be the name of the generic real.

Let (n_k) be a  sequence of C-names for natural numbers.
I will find a sequence (s_k) of names for finite 01-sequences
(s_k of length n_k)
such that C forces:  every old real is in some [s_k].

Let D_k be a dense open set deciding the value of n_k and
containing only conditions of length at least k.
Say, each q in D_k decides that the value of n_k is f_k(q), where
f_k is a function in the ground model defined on D_k.
Each f_k, and also the sequence  (f_k), is in V.



 Now we work in the extension.
 (The point is that even though we now know the actual values of
  n_k, we will play stupid and use the names only, plus the minimal
  amount of information that we need from the generic real.
  This lets us gauge exactly how much information from the
  generic we need.)


 In the extension I will define a sequence (i_k) of natural numbers.
 Let i_k be the minimal i such that c|i is in D_k,
 where c|i = c restricted to i.
 (So i_k is at least k.)
 Now let s_k be the Cohen real restricted to the
 interval [i_k , i_k+n_k].   This is a 01-sequence of length n_k.


I claim that "every old real is in some [s_k]" is forced.
Assume not, so let p force that x is not covered.
Let k be larger than the length of p. So p not in D_k.
Extend p to q so that q is in D_k, q minimal.
Let l be the length of q.
So q forces that i_k is exactly l.    Also q forces that n_k = f_k(q).
Now extend q to q', using the first f_k(q) bits of x.
So q' is stronger than q, and q' forces that s_k is an initial
segment of x.

mg*